Math 172.200
Exam 1 Solutions
February 23, 2010
1. A : Use substitution with u = 1 + sin(2x), du = 2 cos(2x) dx. Then
Z
π
4
0
cos(2x)
1
dx =
1 + sin(2x)
2
Z
2
1
1
du =
u
1
2
ln 2 .
2. D : Use substitution with u = x2 + 1, du = 2x dx. Then
√
Z
3
√
0
5x
1
dx =
2
x2 + 1
4
Z
1
i4
5
5h
√ du = 2u1/2 = 5 .
2
u
1
3. E : The work done is obtained from the following integral:
Z
W =
2
3
2 ln x
dx = 2
x
The substitution used was u = ln x, du =
Z
1
x
ln 3
ln 2
ln 3
u du = u2 ln 2 = (ln 3)2 − (ln 2)2 .
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
2
3
.
cos (x) dx =
cos(x)(1 − sin (x)) dx =
(1 − u2 ) du =
π 0
π 0
π 0
3π
The substitution used was u = sin(x), du = cos(x) dx.
5. E : The limit represents the limit of Riemann sums for the integral
R3
1
1 x2
dx (using n equal length
1 3
subintervals and right-hand end points). Therefore the value of the limit is − x 1 = − 31 − 1 = 23 .
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) Here we must apply integration by parts twice. The first time we let u = e3x and dv = cos x dx,
from which we find du = 3e3x dx and v = sin x. Therefore,
Z
Z
e3x cos x dx = e3x sin x − 3 e3x sin x dx.
To evaluation the integral on the right we use u = e3x and dv = sin x dx (and so du = 3e3x dx and
v = − cos x). Continuing the calculation we find
Z
Z
Z
e3x cos x dx = e3x sin x − 3 −e3x cos x + 3 e3x cos x dx = e3x sin x + 3e3x cos x − 9 e3x cos x dx.
Z
Now we solve for
e3x cos x dx and find that
Z
e3x cos x dx =
1 3x
3
e sin x + e3x cos x + C .
10
10
(b) We use the identity sin A cos B =
Z
1
sin(2x) cos(3x) dx =
2
Z
1
2
sin(A − B) + sin(A + B) with A = 2x and B = 3x. Then
1
1
sin(−x) + sin(5x) dx =
cos x −
cos(5x) + C .
2
10
(c) There
are a couple of ways to do this problem. One was is to first do a substitution, say letting
√
1
w = x and dw = 2√
dx. Thus dx = 2w dw, and so
x
√
Z
e
Z
x
dx = 2
wew dw.
Now apply integration by parts, with u = w and dv = ew dw. Since then du = dw and v = ew , we find
that
Z
Z
w
w
we dw = we − ew dw = wew − ew .
Putting it all together we find.
Z √
√
√ √
e x dx = 2(wew − ew ) = 2 x e x − 2e x + C .
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
2
2
(4 − x ) − (−1) − 1 dx = π
(24 − 10x2 + x4 ) dx
V =π
−2
−2
2
10
1
= π 24x − x3 + x5
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.