Spring 2006 Math 152
Exam 1A: Solutions
c
Mon, 20/Feb
2006,
Art Belmonte
• Integrate term-by-term. Recall that x > 1. Hence
Z
1
dx
x (x − 1) (x + 1)
Z
1
1
−1
=
+ 2 + 2 dx
x
x −1 x +1
= − ln x + 12 ln (x − 1) + 12 ln (x + 1) + C
!
p
x2 − 1
+C
= ln
x
1. (c) We have
1/2
1 Rb
1 R4 2
f ave = b−a
x dx
a f (x) d x = 4−1 1 x − 1
4
√
√
3/2 1
2
= 13
x2 − 1
= 19 · 15 15 − 0 = 53 15.
2
3
1
2. (d) Use integration by parts. First compute an antiderivative,
then apply the FTC.
via the properties of logarithms.
5. (a) If x = sin−1 2t , then x is the angle whose sine (opp / hyp)
is t/2. Draw a right triangle. Then sec x = 1/ cos x (hyp/adj)
u=x
dv = e−2x d x
. Then
• Let
du = d x v = − 12 e−2x
R
R −2x
d x = − 12 xe−2x + 12 e−2x d x
xe
1 −2x
= − 2 xe
− 14 e−2x = − 14 (2x + 1) e−2x .
R
1
• Hence 01 xe−2x d x = − 14 (2x + 1) e−2x 0
−2
= − 34 e−2 − − 14 = 1−3e
.
4
X1A/5
2
t
equals p
5p
Z
25 −
x2 dx
π/2
=
0
0
=
25
2
Z
=
25
2
=
25 π
4
5 cos θ · 5 cos θ dθ
π/2
1 + cos 2θ dθ
π/2
θ + 12 sin 2θ 0
R5p
0
9. (a) This is a trigonometric integral. First compute an
antiderivative, then apply the FTC.
Z
Z
sin 2x 1 − cos2 2x d x
(sin 2x)3 d x =
Z
Z
=
sin 2x d x + (cos 2x)2 (− sin 2x) d x
1
3
= − 12 cos 2x + 12
3 (cos 2x)
4. (d) We’ll integrate the rational function via partial fractions.
• Split the integrand into a sum of partial fractions.
1
x (x − 1) (x + 1)
=
1
=
A
B
C
+
+
x
x −1
x +1
A x2 − 1 + B x2 + x + C x2 − x
0x 2 + 0x + 1
=
( A + B + C) x 2 + (B − C) x + − A
2 1/2
(4 − t )
u = ln (2x)
dv = d x
• Let
. Then
2
1
du = 2x d x = x d x R v = x
R
ln (2x) d x = x ln (2x) − 1 d x
= x ln (2x) − x = x (ln (2x) − 1).
e
Re
• Hence 1 ln (2x) d x = x (ln (2x) − 1) 1
= (e (ln (2e) − 1)) − (ln 2 − 1) =
e (ln 2 + 1 − 1) − ln 2 + 1 = e ln 2 − ln 2 + 1.
25 π.
4
25 − x 2 d x represents the
p
area in the first quadrant under the curve y = 25 − x 2 , part
of the circle x 2 + y 2 = 25 = 52 . This quarter-circular area is
1
1
25
2
2
4 πr = 4 π (5) = 4 π.]
[Alternatively, the integral
.
8. (c) Use integration by parts. First compute an antiderivative,
then apply the FTC.
0
−0=
4 − t2
√
6. (b) When the curves y = x 2 and y =
x intersect, their
√
y-coordinates are equal. Thus x2 = x implies x 4 = x.
Hence 0 = x 4 − x = x x 3 − 1 whence x = 0, 1. Since
q
2
1
1
1
1
2
=
<
=
4
2
4 , we conclude that y = x lies below
√ 16
y = x on [0, 1]. Therefore the area of the region is given
R1√
by 0 x − x 2 d x.
R
R
7. (e) The volume by slicing is V = A (x) d x = y 2 d x
3
R
R3
9 − x 2 d x = 2 03 9 − x 2 d x = 2 9x − 13 x 3 = −3
0
= 2 (27 − 9) − 0 = 36.
3. (b) Use trigonometric substitution. Let x = 5 sin θ . Then
x 0
5
. So
d x = 5 cos θ dθ and we have the table
θ 0 π/2
Z
2
x
R π/2
Therefore, 0 (sin 2x)3 d x = 16 cos3 2x −
= − 16 + 12 − 16 − 12 = 1 − 13 = 23 .
• Equate coefficients of like terms. Thus 1 = −A,
whence A = −1. Next B − C = 0 implies C = B.
Substituting for A and C in A + B + C = 0 yields
2B − 1 = 0, whence B = 12 = C. Therefore,
1
2
π/2
cos 2x 10. (b) Via Hooke’s Law we have F(x) = kx or 12 = 2k,
whence k = 6. The work done is
4
Rb
R4
W = a F (x) d x = 0 6x d x = 3x 2 0 = 48 J.
1
1
−1
1
=
+ 2 + 2 .
x (x − 1) (x + 1)
x
x −1 x +1
1
0
11. When the curves x = 2y and x = 8 − y 2 intersect, their
x-coordinates are equal. Thus 2y = 8 − y 2 implies
0 = y 2 + 2y − 8 = (y + 4) (y − 2) whence y = −4, 2.
Since 2 (0) = 0 < 8 = 8 − 02 , we conclude that x = 2y lies
to the left of x = 8 − y 2 on [−4, 2]. The area of the region is
R2
8 − y 2 − 2y d y, which we now compute.
given by −4
=
2
8y − 13 y 3 − y 2 −4
8
16 − 3 − 4 − −32 +
=
12 −
=
60 −
=
8 + 48 − 64
3
3
72 = 60 − 24
3
− 16
64
3
13. When the curves y = x 2 and y = 2x intersect, their
y-coordinates are equal. Thus x 2 = 2x implies
0 = x 2 − 2x = x (x − 2) whence x = 0, 2. Since
12 = 1 < 2 = 2 (1), we conclude that y = x 2 lies below
y = 2x on [0, 2]. Using washers, the volume swept out by
revolving the region between these curves about the x-axis is
Rb
R2
2
given by a πro2 − πri2 d x = π 0 (2x)2 − x 2 d x, which
we now compute.
Z
2
π
2
− 15 x 5 0
32
32
= π 3 − 5 −0
64π
= 32π 13 − 15 = 32π 5−3
=
15
15
4x 2 − x 4 d x
= π
0
= 36
Here is a picture of the region.
4 x3
3
X1A/P11
8
Here is a figure of the region that is rotated about the x-axis.
6
X1A/13
4
y
2
0
4
−2
−4
−6
3
−6
−4
−2
0
x
2
4
6
8
y
−8
−8
(a) Let 3x = 2 sec θ . Then 3 d x = 2 sec θ tan θ dθ or
d x = 23 sec θ tan θ dθ . Hence (pic at bottom right!)
Z
p
Z
1
9x 2 − 4
dx
=
=
=
or
1
3
2
3
=
1
3
1
3
R
ln |sec θ + tan θ | + C
p
1 3x
9x 2 − 4 ln +
+C
3 2
2
1
5
q
=
=
=
=
1
5
1
5
1
5
Z
x4
1 − x5
p
=
=
=
the left of x = y 1/3 on [0, 1]. Using cylindrical shells, the
volume swept out by revolving the region between these
R
curves about the x-axis is given by cd 2πr w d y
R1
= 2π 0 y y 1/3 − y 2 d y, which we now compute.
2 d x
1
1 − u2
2
Z
1
2π
y 4/3 − y 3 d y
sin−1
=
0
du
=
sin−1 u
(c) Use integration by parts.
u = tan−1 x
dv = x d x
Let
.
1
v = 12 x 2
du = 1+x
2 dx
x tan−1 x d x
1
x
14. When the curves x = y 2 and x = y 1/3 intersect, their
x-coordinates are equal. Thus y 2 = y 1/3 implies
0 = y 6 − y = y y 5 − 1 whence y = 0, 1. Since
1/3
2
1
1
1
1
=
<
=
, we conclude that x = y 2 lies to
8
64
2
8
du = x 4 d x. Thus
Z
x4
p
dx
1 − x 10
or
0
sec θ dθ
(b) Let u = x 5 . Then du = 5x 4 d x or
Z
0
sec θ tan θ dθ
2 tan θ
p
ln 3x + 9x 2 − 4 + K via log properties.
Z
2
1
+C
x 5 + C.
=
1
− 14 y 4 0
3
1
2π 7 − 4 − 0
5π
2π 12−7
=
28
14
2π
3 y 7/3
7
Here is a figure of the region that is rotated about the x-axis.
X1A/14
X1A/12
1.5
Then
1
Z
1 2 −1
1
x2
dx
x tan x −
2
2
1 + x2
Z
1 2 −1
1
1
dx
x tan x −
1−
2
2
1 + x2
1 2 −1
1
1
x tan x − x + tan−1 x + C
2
2
2
y
12.
0.5
3x
(9x2 −4)1/2
0
x
0
0.5
x
x 2 + 1 tan−1 x − x
+ C.
2
2
1
1.5
2