Math 172.503
Exam 1 Solutions
February 23, 2010
1. A : Use substitution with u = 1 + sin(2x), du = 2 cos(2x) dx. Then
Z
Z π4
cos(2x)
1 21
dx =
du = 12 ln 2 .
1 + sin(2x)
2 1 u
0
2. D : Use substitution with u = x2 + 1, du = 2x dx. Then
√
Z
0
3
5x
1
√
dx =
2
2
x +1
Z
1
4
i4
5
5h
√ du = 2u1/2 = 5 .
2
u
1
3. E : The work done is obtained from the following integral:
Z 3
Z ln 3
ln 3
2 ln x
W =
dx = 2
u du = u2 ln 2 = (ln 3)2 − (ln 2)2 .
x
2
ln 2
The substitution used was u = ln x, du =
1
x
dx.
4. C : Since the length of the interval 0 ≤ x ≤ π2 is π2 , the average value is obtained from the following
integral:
Z π2
Z π2
Z
2
2
2 1
4
.
cos3 (x) dx =
cos(x)(1 − sin2 (x)) dx =
(1 − u2 ) du =
π 0
π 0
π 0
3π
The substitution used was u = sin(x), du = cos(x) dx.
Rx
5. B : Let G(x) = 3 √t21+3t dt. Then by the Fundamental Theorem of Calculus, G0 (x) = √x21+3x . Since
F (x) = G(2x), we see that F 0 (x) = 2G0 (2x) by the Chain Rule. Therefore,
F 0 ( 12 ) = 2G0 (2 · 21 ) = 2G0 (1) = √
2
= 1.
1+3
6. To find the intersection points of the curves, we solve 1 − y 2 = y 4 − y 2 − 15, which quickly simplifies to
y 4 = 16. The real solutions of this equation are y = ±2. Therefore, the area of the region in question
is
Z 2
Z 2
256
2
4
2
.
A=
(1 − y ) − (y − y − 15) dy =
(16 − y 4 ) dy =
5
−2
−2
7. (a) For this integral we substitute u = x3 − 1, du = 3x2 dx. Thus also x3 = u + 1 and x2 dx = 31 du.
Therefore,
Z
Z
Z 1
1
1
x5
u+1
1
dx =
du =
1+
du = u + ln |u| + C = 13 (x3 − 1) + ln |x3 − 1| + C .
x3 − 1
3
u
3
u
3
(b) Here we use integration by parts. Take u = x and dv = cos(3x) dx. Then du = dx and v =
Using the integration by parts formula we see that
Z
Z
x
1
x cos(3x) dx = sin(3x) −
sin(3x) dx = x3 sin(3x) + 91 cos(3x) + C .
3
3
(c) We use the identity sin2 (x) = 12 (1 − cos(2x)). Then
Z
Z
1
1
1
2
sin (x) dx =
(1 − cos(2x)) dx = (x − sin(2x)) + C =
2
2
2
x
2
−
1
4
sin(2x) + C .
1
3
sin(3x).
8. Since 4 − x2 = 0 when x = ±2, we see that the region we are rotating sits on the interval −2 ≤ x ≤ 2.
Using the washer method, we see that the volume obtained by rotating about the line y = −1 is
Z 2
Z 2h
i
2
(24 − 10x2 + x4 ) dx
V =π
(4 − x2 ) − (−1) − 12 dx = π
−2
−2
2
10 3 1 5
= π 24x − x + x
3
5
−2
80 32
80 32
= π 48 −
+
− π −48 +
−
3
5
3
5
=
832π
.
15
9. The first point of intersection between y = sin(2x) and y = sin(x) to the right of the origin is at x =
On the interval 0 ≤ x ≤ π3 we have sin(2x) ≥ sin(x).
(a) Rotating about the x-axis we use the washer method and find
π
3
Z
V =π
2
sin (2x) − sin2 (x) dx
0
(b) Rotating about the y-axis we use cylindrical shells and find
Z
V = 2π
0
π
3
x sin(2x) − sin(x) dx
π
3.