MTH131 Applied Calculus – Spring 2016
Lab 8 – SOLUTIONS
1. (a) Let u = x3 + 4, du = 3x2 dx. Subbing in we obtain
Z
Z
p
3x2 x3 + 4 dx =
√
u du,
2 3/2
u + C,
3
2 3
(x + 4)3/2 + C.
3
=
=
(b) Let u = 5 − x, du = −dx. Subbing in we obtain
1
(−dx),
5−x
Z
1
= −
du,
u
= − ln |u| + C,
1
dx = −
5−x
Z
Z
= − ln |5 − x| + C.
(c) Let u = x3 − 1, du = 3x2 dx. Subbing in we obtain
Z
2
3
4
x (x − 1) dx =
=
=
=
=
1
3x2 (x3 − 1)4 dx
3
Z
1
u4 du
3
1 1 5
u +C
3 5
u5
+C
15
(x3 − 1)5
+C
15
Z
(d) Let u = x3 − 3x, du = (3x2 − 3) dx = 3(x2 − 1) dx. Subbing in we obtain
Z
2
3
4
(x − 1)(x − 3x) dx =
=
=
=
=
1
3(x2 − 1)(x3 − 3x)4 dx
3
Z
1
u4 du
3
1 1 5
u +C
3 5
u5
+C
15
(x3 − 3x)5
+C
15
Z
(e) Let u = 1 + 4x, du = 4 dx. Subbing in we obtain
Z
5
dx =
1 + 4x
=
=
=
5
4 dx
4 1 + 4x
Z
5 du
4
u
5
(ln |u| + C) ,
4
5
ln |1 + 4x| + C,
4
Z
(f) Let u = x2 − 1, du = 2x dx. Subbing in we obtain
Z
xe
x2 −1
1
2
2xex −1 dx,
2
Z
1
eu du,
2
1 u
e + C,
2
1 x2 −1
e
+ C.
2
Z
dx =
=
=
=
(g) Let u = x + 1, du = dx. Note also that x − 2 = u − 3. Subbing in we obtain
Z
12(x − 2)(x + 1)5 dx =
Z
12(u − 3)u5 du,
Z
(u6 − 3u5 ) du,
1 7 1 6
u − u + C,
7
2
= 12
= 12
=
12
(x + 1)7 − 6(x + 1)6 + C.
7
(h) Let u = x + 1, du = dx. Subbing in we obtain
Z
u−1
du
u
Z 1
=
1−
du
u
= u − ln |u| + C
x
dx =
x+1
Z
= x + 1 − ln |x + 1| + C, or
= x − ln |x + 1| + C, since we can incorporate the 1 into the constant C
(i) Let u = e2x + 1, du = 2e2x dx. Subbing in we obtain
Z
e2x
dx =
e2x + 1
=
=
=
1 2e2x dx
,
2
e2x + 1
Z
1 du
,
2
u
1
ln |u| + C,
2
1
ln |e2x + 1| + C.
2
Z
(j) Let u = e2x + 1, du = 2e2x dx. Subbing in we obtain
Z
e4x
dx =
e2x + 1
1 2e2x e2x dx
,
2
e2x + 1
Z
e2x
1
2e2x dx,
2 e2x + 1
Z
1 u−1
du,
2 u
Z
1
1
1−
du,
2
u
1
(u − ln |u|) + C,
2
1 2x
e + 1 − ln |e2x + 1| + C, or
2
1 2x
e − ln |e2x + 1| + C.
2
Z
=
=
=
=
=
=
2. (a) Let u = x3 + 4, du = 3x2 dx. x = 0 ⇒ u = 4 and x = 1 ⇒ u = 5. Subbing in we obtain
Z 1
3x
2
p
x3
+ 4 dx =
Z 5
√
u du,
4
0
2 3/2 5
u ,
3
4
2
2 3/2
(5 ) − (43/2 ),
3√
3
2 125 16
−
≈ 2.12.
3
3
=
=
=
(b) Let u = x2 − 3, du = 2x dx. x = 2 ⇒ u = 1 and x = 3 ⇒ u = 6. Subbing in we obtain
Z 3
2
x
dx =
x2 − 3
=
=
=
1 3 2x
dx,
2 2 x2 − 3
Z
1 6 du
,
2 1 u
6
1
ln |x| ,
2
1
ln 6
1
(ln 6 − ln 1) =
≈ 0.896.
2
2
Z
(c) Let u = x3 + 73, du = 3x2 dx. x = 2 ⇒ u = 81 and x = 3 ⇒ u = 100. Subbing in we
obtain
x2
√
dx =
x3 + 73
Z 3
2
1 3 3x2 dx
√
,
3 2
x3 + 73
Z
1 100 −1/2
u
du,
3 81
100
1
(2u1/2 ) ,
3
81
2
1/2
(100 − 811/2 ),
3
2
2
(10 − 9) = .
3
3
Z
=
=
=
=
(d) Let u = x3 , du = 3x2 dx. x = 0 ⇒ u = 0 and x = 2 ⇒ u = 8. Subbing in we obtain
Z 2
4 2 2 x3
3x e dx,
3 0
Z
4 8 u
e du,
3 0
4 u 8
e ,
3 0
4 8
(e − 1) ≈ 3973.28.
3
Z
3
4x2 ex dx =
0
=
=
=
(e) Let u = ln x, du =
dx
x .
x = e ⇒ u = 1 and x = e2 ⇒ u = 2. Subbing in we obtain
Z e2
1
dx =
x ln x
Z e2
1 dx
ln x x
Z 2
du
,
=
1 u
= ln |x| |21 ,
e
,
e
= ln 2 − ln 1 = ln 2 ≈ 0.693
3. (a) When x is in the range [0, 2], ex is the “upper” function, so the area between the curves
is given by
Z 2
0
(ex − e−x ) dx = (ex + e−x )|20 = (e2 + e−2 ) − (e0 + e−0 ) = e2 + e−2 − 2 ≈ 5.52
(b) We proceed as above with different limits of integration
Z ln 2
0
1
1
2
(ex − e−x ) dx = (ex + e−x )|ln
= (eln 2 + e−ln 2 ) − (e0 + e−0 ) = (2 + ) − 2 = .
0
2
2
(c) We need to evaluate the integral
Z 1
0
x
e −
√
√
xe
x3
Z 1
dx =
0
x
e dx −
Z 1
√
0
√
xe
x3
dx
We will do this in two steps, evaluating each of the integrals on the right hand side
separately. The first one is straight forward:
Z 1
ex |10
ex dx =
0
= e1 − e0 = e − 1.
√
To evaluate the second one, let u = x3 = x3/2 , du = 32 x1/2 . x = 0 ⇒ u = 0 and
x = 1 ⇒ u = 1. Subbing in we obtain
Z 1
√
√
xe
x3
2 1 3 √ √ x3
xe
dx,
3 0 2
Z 1
2
eu du,
3 0
2 u 1
e 3 0
2
(e − 1)
3
Z
dx =
0
=
=
=
Therefore the area between the two curves is
Z 1
ex −
√
√
xe
x3
0
2
dx = (e − 1) − (e − 1)
3
1
=
(e − 1) ≈ 0.573
3
4. The general formula for the average value of a function f (x) over an interval [a, b] is
1
b−a
Z b
f (x) dx.
a
We’ve already calculated the value of the first four integrals in Problem 5 of the previous lab,
and the last two integrals have been calculated above in Problem 2. It’s now just a matter of
dividing by the (b − a) to get the average value.
(a) f (x) = 12 − 3x2 on the interval [1, 2]
1
2−1
Z 2
2
2
1
(12x − x3 )
1
1
(12 − 3x ) dx =
1
= 5
(b) f (x) =
1
on the interval [1, 4]
x3
1
4−1
1 −1 4
dx
=
x3
3 2x2 1
1 15
5
=
=
3 32
32
Z 4
1
1
(c) f (x) =
1
on the interval [1, 4]
x
1
4−1
Z 4
1
x
1
4
1
ln |x|
3
1
ln 4
3
dx =
=
(d) f (x) = 5ex on the interval [−1, 2]
1
2 − (−1)
Z 2
2
1
(5ex )
3
1
5e2 − 5e−1
3
x
5e dx =
−1
=
(e) f (x) = √
x2
on the interval [2, 3] (using the substitution u = x3 + 73)
x3 + 73
1
3−2
Z 3
2
√
x2
dx =
x3 + 73
1
1
100
1
(2u1/2 )
3
81
2
3
=
3
(f) f (x) = 4x2 ex on the interval [0, 2] (using the substitution u = x3 )
1
2−0
Z 2
3
4x2 ex dx =
0
=
1
2
4 u 8
e ,
3 0
2 8
(e − 1) ≈ 1986.64.
3