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446 • Improper Integrals: Basic Concepts
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f(
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
In this picture, the dotted and dashed curves are typical of what the graphs
of y = 1/x
p
look like for p < 1 and p > 1, respectively. The solid curve is
y = 1/x, which isn’t quite close enough to the y-axis for
R
1
0
1/x dx to converge,
nor close enough to the x-axis for
R
∞
1
1/x dx to converge. On the other hand,
for any p < 1, the dotted curve is close enough to the y-axis for
R
1
0
1/x
p
dx to
converge. The situation is reversed when you look for proximity to the x-axis:
there we need to look at the dashed curve, representing y = 1/x
p
for p > 1,
to get close enough for
R
∞
1
1/x
p
dx to converge.
Notice that since 1.0000001 > 1, the integral
Z
∞
1
1
x
1.0000001
dx
converges even though
R
∞
1
1/x dx diverges! Just nudging the power of x from
1 up to 1.0000001 was enough to make all the difference. This shows how
incredibly delicate this whole issue of convergence and divergence really is.
Now, let’s prove the p-test. Luckily, this is just a computation using the
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
formulas from Section 20.1 above. First, consider
Z
∞
a
1
x
p
dx
for some constant a > 0. If p = 1, the integrand becomes 1/x, and we’ve
already seen that the integral diverges in that case. Otherwise, we have
Z
∞
a
1
x
p
dx = lim
N→∞
Z
N
a
x
−p
dx = lim
N→∞
1
1 − p
x
1−p
N
a
=
1
1 − p
lim
N→∞
N
1−p
− a
1−p
.
Now if the limit
lim
N→∞
N
1−p
Section 20.6: The Absolute Convergence Test • 447
exists, then the whole integral
R
∞
a
1/x
p
dx converges. If on the other hand
the limit doesn’t exist, the integral diverges. So, write the above limit as
lim
N→∞
1
N
p−1
.
If p > 1, then p−1 > 0, so N
p−1
gets very large as N gets large. Its reciprocal
becomes very small, and so the limit is therefore 0 and our original integral
converges. On the other hand, if p < 1, then 1 − p > 0, so N
1−p
gets very
large and the limit blows up to ∞, proving that the original integral diverges.
This proves one half of the p-test. The proof of the other half is almost the
same; you just have to use ε → 0
+
instead of N → ∞. I’ll leave the details
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
to you.
20.6 The Absolute Convergence Test
One of the assumptions in the comparison test is that the functions f and g are
always nonnegative. What if you want to investigate the behavior of a function
which is sometimes negative? Well, if the function is always negative, you
could just pull out a minus sign and reduce it to the case of a positive function.
We’ll see an example of this in the next chapter. On the other hand, if the
function keeps oscillating between positive and negative values throughout
the region of integration, you can appeal to the absolute convergence test.
Here’s what it says:
if
Z
b
a
|f(x)|dx converges, then so does
Z
b
a
f(x) dx.
This also works on infinite regions of integration (such as [a, ∞) instead of
[a, b]). Watch out: if the absolute-value version of the original integral di-
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
verges, then the original integral could still converge! Such examples are
pretty cool but they’re beyond the scope of this book. On the other hand,
we’ll see something similar when we look at alternating series in Section 23.7
of Chapter 23.
Why is the above test useful? Well, for one thing, |f(x)| is always non-
negative, so you can use the comparison test on improper integrals involving
it. For example, consider the improper integral
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
∞
1
sin(x)
x
2
dx.
The integrand sin(x)/x
2
oscillates between positive and negative values as x
gets larger and larger without bound. So we can’t use the comparison test
∗
or the limit comparison test yet. Let’s try the absolute convergence test first.
∗
Direct comparison won’t work, since the integrals
R
N
1
sin(x)/x
2
dx are not increasing
in N as N gets bigger. The idea of the argument at the end of Section 20.3 above fails
since it depends on the integrals getting bigger and bigger without bumping into the ceiling
provided by the integral of g.
448 • Improper Integrals: Basic Concepts
We need to consider this integral instead:
Z
∞
1
sin(x)
x
2
dx.
This can be rewritten as
Z
∞
1
|sin(x)|
x
2
dx
since x
2
can’t be negative. Now we can use the comparison test. You see,
|sin(x)| ≤ 1 for all x, so it follows that
|sin(x)|
x
2
≤
1
x
2
for all x. The comparison test says that
Z
∞
1
|sin(x)|
x
2
dx ≤
Z
∞
1
1
x
2
dx.
Since the right-hand integral converges by the p-test, so does the left-hand
integral. Finally, we can use the absolute convergence test to say that
Z
∞
1
|sin(x)|
x
2
dx converges, so
Z
∞
1
sin(x)
x
2
dx also converges.
It’s a little subtle, but you really do need to use those absolute values.
Here’s another example:
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
∞
0
cos(x) dx.
The integrand cos(x) oscillates between positive and negative values, so maybe
we should look at the absolute value “version” of the integral:
Z
∞
0
|cos(x)|dx.
Unfortunately, there’s not a hope in hell that this new integral converges.
To see why, draw the graph of y = |cos(x)| right now and you’ll see that
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
it’s a series of identical humps, one after the other. There’s no way you can
add up the areas of infinitely many identical humps and get a finite value.
So the absolute value version diverges. This means that we cannot use the
absolute convergence test! The only time that this test can be used is when
the absolute value version of the integral converges.
We have learned nothing from these shenanigans: we are back to square
one. We don’t know whether our original integral converges or diverges. So,
let’s try using the definition of the improper integral with problem spot at ∞:
Z
∞
0
cos(x) dx = lim
N→∞
Z
N
0
cos(x) dx = lim
N→∞
sin(x)
N
0
= lim
N→∞
(sin(N) − sin(0)) = lim
N→∞
sin(N).
Section 20.6: The Absolute Convergence Test • 449
This last limit doesn’t exist, since sin(N) keeps on oscillating between −1 and
1, never making up its mind even as N becomes larger and larger without
bound. So our original integral
R
∞
0
cos(x) diverges, not because it goes to ∞
or −∞, but because it oscillates too much.
Oscillating integrals like this are extremely tricky to deal with. If you’re
lucky, you can use the formal definition as we did above. Most of the time this
doesn’t work. Many mathematicians have spent a whole lot of time trying to
understand what’s going on. For the moment, just bear the above example in
mind. We’ll have more than enough to deal with in the next chapter, where we
return to the tests and see how to solve problems involving improper integrals.
Before we do this, let’s take a quick look at why the absolute convergence
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
test works. Suppose we know that
Z
b
a
|f(x)|dx
converges. Now comes a nice trick: set g(x) = |f(x)| + f(x) for all x in [a, b]
where f is defined. Then g has two important properties: first, g(x) ≥ 0, and
second, g(x) ≤ 2|f(x)|. (In both cases, x is any number in [a, b] which is in
the domain of f.) In fact, if you think about it, you can see that g(x) actually
equals 2f(x) whenever f(x) ≥ 0, and that g(x) actually equals 0 whenever
f(x) < 0. Try to show that the two important properties follow from this.
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Anyway, we can now use the comparison test on g:
0 ≤
Z
b
a
g(x) dx ≤ 2
Z
b
a
|f(x)|dx < ∞.
The conclusion is that
Z
b
a
g(x) dx
converges as well. So what? Well, notice that f(x) = g(x) − |f(x)|, so
Z
b
a
f(x) dx =
Z
b
a
g(x) dx −
Z
b
a
|f(x)|dx.
Both integrals on the right converge—the first because we just showed it, and
the second because we are assuming it—so the left-hand side converges as
well.
C h a p t e r 21
Improper Integrals: How to Solve Problems
Let’s get practical and look at a lot of examples of improper integrals. As
we go along, we’ll summarize the main methods. In the previous chapter,
we introduced some tests that will turn out to be really useful. To use them
effectively, you have to understand how some common functions behave, espe-
cially near 0 and near ∞. By “common functions,” I mean our usual suspects:
polynomials, trig functions, exponentials, and logarithms. So, here’s the game
plan for this chapter:
• what to do when you first see an improper integral, including how to deal
with multiple problem spots and functions which aren’t always positive;
• summary of the comparison test, limit comparison test, and p-test;
• the behavior of common functions near ∞ and −∞;
• the behavior of common functions near 0; and
• how to handle problem spots at finite values other than 0.
21.1 How to Get Started
OK, so you have an improper integral,
R
b
a
f(x) dx. (We’ll always assume that
f is continuous or has finitely many discontinuities.) You know that your
integral is improper because the integrand f has at least one problem spot in
[a, b]. Problem spots occur at blow-up points of f , like vertical asymptotes,
and also at ∞ and −∞, if applicable. For example, the integral
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
∞
−∞
1
x
2
− 1
dx
has problem spots at ∞ and −∞ (since these are always problem spots if they
are involved), and also at x = 1 and x = −1 (since the integrand is undefined
there).
As we said in Section 20.1.2 of the previous chapter, it makes sense to
concentrate on one problem spot at a time. Also, we’d like to arrange matters
so that the integrand is always positive, at least when x is near the problem
452 • Improper Integrals: How to Solve Problems
spot. So, our first task is to split up the integral as appropriate, and our
second task will be to deal with what happens if f is sometimes negative.
21.1.1 Splitting up the integral
Here’s the basic plan of attack:
1. Identify all the problem spots in the region [a, b].
2. Split up the integral into enough pieces so that each new piece has
at most one problem spot, which occurs at one of the endpoints of the
integral.
3. Look at each piece individually. If any one piece diverges, so
does the whole thing. The only way the original improper integral
can converge is if each piece converges.
How do you split up an integral into the correct pieces? If there’s only one
problem spot which is either at a or b, there’s nothing to do. On the other
hand, consider this classic example: does
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
∞
0
1
√
x + x
2
dx
converge or diverge? Well, the integrand has a vertical asymptote at x = 0,
and ∞ is always a problem spot, so we have problem spots at the endpoints
0 and ∞. That’s two problem spots, and we can only cope with one per
integral. So pick your favorite number between 0 and ∞—mine’s 5 at the
moment—and split up the integral into two pieces:
Z
5
0
1
√
x + x
2
dx and
Z
∞
5
1
√
x + x
2
dx.
We’ll finish off this example in Section 21.4.1 below. For the moment, notice
that both of these integrals have one problem spot only, and in both cases
the problem spot is at the left or right endpoint of the region of integration.
It doesn’t matter where you split the integral up—this point was discussed
at length at the end of Section 20.1.1 in the previous chapter. You should
also check out the example on page 436 involving an improper integral which
needs to be split up into five different pieces.
As for the third step above, we’ll spend the rest of this chapter looking at
how to deal with individual pieces with only one problem spot occurring at
one of the endpoints. The key point is that all the pieces have to converge
in order for the whole integral to converge. So if you break up an improper
integral into five pieces, for example, and you look at one of the pieces and find
that it diverges, don’t waste your time looking at the other four pieces—you
already know that the whole integral diverges.
Here’s an important case: what if there are no problem spots? That is,
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−2
0
y = e
−x
2
1
2
e
−1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
suppose you have an integral
R
b
a
f(x) dx such that the region [a, b] of inte-
gration is bounded (so no ∞ or −∞), and f is bounded on all of the closed
interval [a, b]. Well, then, as we saw in Section 20.1 of the previous chapter, f
has no problem spots, so we know that the integral
R
b
a
f(x) dx converges. In
Section 21.1.2: How to deal with negative function values • 453
summary, if there are no problem spots, the integral automatically
converges! So, for example,
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
100
0
ln(x + 1)
x
4
+ x
2
+ 1
dx
converges since the integrand is bounded on the bounded region [0, 100]—that
is, there are no problem spots. Don’t get suckered into using any fancy tests
in an example like this.
21.1.2 How to deal with negative function values
If f(x) takes on negative values for some x in [a, b], which often happens when
trig functions or logs are present, you need to take special care. Luckily you
can often reduce matters to integrals with only positive integrands. Here are
three ways to deal with negative function values:
1. If the integrand f(x) is both positive and negative as x ranges over [a, b],
you should consider trying the absolute convergence test. As we saw in
Section 20.6 of the previous chapter, this says that
if
Z
b
a
|f(x)|dx converges, then so does
Z
b
a
f(x) dx.
This test is particularly useful for investigating improper integrals in-
volving trig functions when the region of integration is unbounded. The
example
Z
∞
1
sin(x)
x
2
dx
from Section 20.6 of the previous chapter is of this type. Recall that the
way to start is to consider the absolute version of the integral, namely,
Z
∞
1
|sin(x)|
x
2
dx;
you don’t need absolute values around the denominator since it’s always
positive. Then show that this new integral converges (see page 448 for
the details) and conclude that the original integral converges as well, by
the absolute convergence test. In general, don’t forget this important
point: the absolute convergence test only helps you show that an integral
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
converges. That is, you cannot use the absolute convergence test
to show that an integral diverges!
2. Suppose that the integrand f(x) is always negative (or zero) on [a, b].
That is, f(x) ≤ 0 on [a, b]. If this is true, you can write
Z
b
a
f(x) dx = −
Z
b
a
(−f(x)) dx.
So what? Well, −f(x) is now always nonnegative, so you can use the
comparison test or the p-test to see whether
R
b
a
(−f(x)) dx converges
454 • Improper Integrals: How to Solve Problems
or diverges. Of course, if this integral converges, so does
R
b
a
f(x) dx,
and similarly if
R
b
a
(−f(x)) dx diverges, so does
R
b
a
f(x) dx. Here’s an
example: consider
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
Z
1/2
0
1
x
2
ln(x)
dx.
There’s certainly a problem spot at x = 0. The thing to realize is that
ln(x) is actually negative for x between 0 and 1, so it’s a good idea to
start out by writing
Z
1/2
0
1
x
2
ln(x)
dx = −
Z
1/2
0
−1
x
2
ln(x)
dx.
Actually, since ln(x) is the negative part here, you can replace −ln(x)
by |ln(x)| as follows:
Z
1/2
0
1
x
2
ln(x)
dx = −
Z
1/2
0
1
x
2
|ln(x)|
dx.
Now we can just worry about
Z
1/2
0
1
x
2
|ln(x)|
dx.
Unfortunately you’ll have to wait until page 474 to see that this last
integral diverges. The conclusion will then be that the original integral
diverges as well. Note that the absolute convergence test doesn’t work in
this case, since that test can only be used to show an improper integral
converges.
3. If neither of the previous two cases seems to apply, you may be able to
use the formal definition of the improper integral to see what’s going on.
An example of this is
Z
∞
0
cos(x) dx,
which we looked at on page 448.
This is not the end of the story. There are slightly freaky improper integrals
which converge, but which are not absolutely convergent.
∗
These sorts of im-
proper integrals seem to come up quite often in actual physics and engineering
applications, but they are beyond the scope of this book. So, it’s time to go
back and review the integral tests.
21.2 Summary of Integral Tests
The most valuable tools you have at your disposal are the comparison test,
the limit comparison test, and the p-test. We looked at these tests from a
∗
For example,
R
∞
1
sin(x)/x dx converges but
R
∞
1
|sin(x)|/x dx diverges. Kudos to you if
you can work out why either of these assertions are true.
Section 21.2: Summary of Integral Tests • 455
theoretical point of view in the previous chapter; here are the statements
once again, for reference. In all the tests below, the integrand f (x) is
assumed to be positive on the region of integration.
• Comparison test, divergence version: if you think that
R
b
a
f(x) dx
diverges, find a smaller function whose integral also diverges. That is,
find a nonnegative function g such that f(x) ≥ g(x) on (a, b), and such
that
R
b
a
g(x) dx diverges. Then
Z
b
a
f(x) dx ≥
Z
b
a
g(x) dx = ∞,
so
R
b
a
f(x) dx diverges.
• Comparison test, convergence version: if you think that
R
b
a
f(x) dx
converges, find a larger function whose integral also converges. That is,
find a function g such that f(x) ≤ g(x) for all x in (a, b), and such that
R
b
a
g(x) dx converges. Then
Z
b
a
f(x) dx ≤
Z
b
a
g(x) dx < ∞,
so
R
b
a
f(x) dx also converges.
Beware of the whoop-di-doo case! This was discussed in Section 20.3 of the
previous chapter, and arises if you get the above inequalities the wrong way
around. The comparison test just doesn’t work if you screw up the direction
of the inequalities.
As an alternative to the comparison test, there is the limit comparison
test. This is useful when you can find a function which behaves just like the
integrand near the problem spot. In Section 20.4.1 in the previous chapter,
we made the following definition:
f(x) ∼ g(x) as x → a means the same thing as lim
x→a
f(x)
g(x)
= 1.
The definition also applies if you replace both instances of x → a by x → ∞
(or x → −∞). In any case, if your integrand f is really nasty and you can
find a nicer function g such that f (x) ∼ g(x) as x approaches the problem
spot, you’re in business! That’s because the limit comparison test says that
whatever goes for g also goes for f. More precisely, here are two versions of
the test depending on whether the problem spot is infinite or finite:
• Limit comparison test, ∞ version: find a simpler nonnegative func-
tion g with no problem spots in [a, ∞), such that f (x) ∼ g(x) as x → ∞.
Then
– if
R
∞
a
g(x) dx converges, so does
R
∞
a
f(x) dx; whereas
– if
R
∞
a
g(x) dx diverges, so does
R
∞
a
f(x) dx.
Of course, you can change the region [a, ∞) into (−∞, b] and everything still
works. There’s also a version which applies when the problem spot is at some
finite value a, which is at the left endpoint of the region of integration: