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Evaluating Definite Integrals by Substitution
Solutions To Selected Problems
Calculus 9th Edition Anton, Bivens, Davis
Matthew Staley
November 15, 2011
1. Express the integral in terms of the variable u, but do not evaluate it.
Z
3
(2x − 1)3 dx;
(a)
u = 2x − 1
1
du = 2 dx
1
du = dx
2
x = 1 → u = 2(1) − 1 = 1
x = 3 → u = 2(3) − 1 = 5
Z
3
1
Z
1
(2x − 1) dx =
2
3
4
(b)
√
3x 25 − x2 dx;
Z
5
u3 du
0
u = 25 − x2
0
du = −2x dx
−
1
du = x dx
2
x = 0 → u = 25 − 02 = 25
x = 4 → u = 25 − 16 = 9
Z
4
√
3x 25 −
0
x2
1
dx = −
2
1
9
√
3
3 u du =
2
25
Z
Z
9
25
√
u du
2. Evaluate the definite integral by making an appropriate u-substitution.
(a)
Z
π/2
4 sin(x/2) dx;
0
x
2
1
du = dx
2
2 du = dx
let u =
x = 0 → u = 0/2 = 0
x = π/2 → u = π/4
Z
π/2
Z
π/4
4 sin(x/2) dx = 2(4)
sin(u) du
0
h
i
= 8 − cos(u)|π/4
0
0
= −8[cos(π/4) − cos(0)]
"√
#
√
2
= −8
−1 = 8−4 2
2
(b)
Z
−1
−2
x
dx;
(x2 + 2)2
let u = x2 + 2
du = 2x dx
1
du = dx
2
x = −2 → u = 4 + 2 = 6
x = −1 → u = 1 + 2 = 3
2
Z
−1
−2
3
Z
x
1
dx =
2
2
(x + 2)
2
6
=−
1
1
du = −
3
u
2
Z
6
u−3 du
3
!
u 1 1
1
=
−
−2 3
4 62 32
−2 6
1
2
1 1
1
1 1−4
=
−
=
4 36 9
4
36
=
−3
1
1
=−
= −
4(36)
4(12)
48
3. Evaluate the definite integral by expressing it in terms of u and evaluating the
resulting integral using a formula from geometry.
(a)
Z
5/3
√
25 − 9x2 dx;
u = 3x
−5/3
du = 3 dx
1
du = dx
3
x = −5/3 → u = −5
x = 5/3 → u = 5
Z
5/3
−5/3
√
Z
p
1 5√
2
dx =
25 − (3x) dx =
25 − u2 du
3
−5/3
−5
Z
25 −
9x2
5/3
√
Now let y = 25 − u2 . Note that this is the same as y 2 +u2 = 25, which is
the top half of a circle centered at the origin with a radius of 5. The area
25
of half of the circle is 12 πr2 = 25
π. So the total integral is 13 25
π =
π
2
2
6
3
4. Evaluate the integrals by any method.
Z
3
√
(a)
1
x+2
dx;
x2 + 4x + 7
let u = x2 + 4x + 7
du = 2x + 4 dx
1
du = (x + 2) du
2
x = 1 → u = 1 + 4 + 7 = 12
x = 3 → u = 32 + 4(3) + 7 = 28
Z
1
(b)
Z
3
Z
x+2
1 28 du
1 1/2 28 √
√
dx =
2u 12
=
2 12
2
u
x2 + 4x + 7
√
√
√
√
√ √
= 28 − 12 = 2 7 − 2 3 = 2
7− 3
π/4
4 sin(x) cos(x) dx let u = sin(x)
0
du = cos(x) dx
x = 0 → u = sin(0) = 0
√
2
x = π/4 → u = sin(π/4) =
2
Z
√
π/4
Z
4 sin(x) cos(x) dx = 4
0
2/2
u du = 4
0
√2/2 !
1 2 u
2 0

√ !2
2
2
2
= 2
−0 =2
= 1
2
4

4
Z
1
√
(c)
0
y2
dy
4 − 3y
let u = 4 − 3y
du = −3 dy
−
1
du = dy
3
y=0→u=4
y=1→u=1
Need to solve for y, then find y 2
u − 4 = −3y
1
− (u − 4) = y
3
1
y 2 = (u2 − 8u + 16)
9
Z
0
1
y2
1
√
dy = −
3
4 − 3y
1 1
(u2
9
Z
4
4
− 8u + 16)
√
du
u
8u
16
u2
√ − √ + √ du
u
u
u
1
=
27
Z
1
=
27
Z
1
=
27
"
1
=
27
2 5/2 16 3/2
2
16
1/2
(4) − (4) + 32(4)
−
(1) − (1) + 32(1)
5
3
5
3
1
=
27
2 5 24 3
2 24
5
5
(2) − (2) + 2 (2) −
−
+2
5
3
5
3
1
4
u3/2 − 8u1/2 + 16u−1/2 du
1
2 5/2
u −8
5
5
2 3/2
u
3
4 #
+ 16(2u1/2 )
1
1
=
27
26 (3) − 27 (5) + 26 (15)
15
−
2(3) − 24 (5) + 25 (15)
15
=
1 1 6
·
3(2 − 2) − 5(27 − 24 ) + 15(26 − 25 )
27 15
=
1
[3(62) − 5(112) + 15(32)]
405
=
106
1
[186 − 560 − 480] =
405
405
6