Spring 2005 Math 152
Exam 1A: Solutions
c
Mon, 21/Feb
2005,
Art Belmonte
7. (b) Use the trig identity sin2 x + cos2 x = 1.
Z
Z sin3 x cos2 x d x =
1 − cos2 x cos2 x sin x d x
Z =
cos4 x − cos2 x (− sin x) d x
For specificity, lengths are in centimeters unless stated otherwise.
Graphs appear at the bottom of the right column to conserve space.
1. (b) We have
R √π
1 Rb
√1
f
(x)
d
x
=
x sin x 2 d x
f ave = b−a
a
0
π−0
√π = − 2√1 π cos x 2 0 = 12 · √1π − − 12 · √1π =
=
√1 .
π
A
Z
2
=
1 − x dx +
x2 − 1 dx
0
1
1
2
x − 13 x 3 0 + 13 x 3 − x 1
2 −
(0) + 83 − 2 − 13 − 1
3
=
4
3
=
=
• Now x 2 = 2x implies x 2 − 2x = 0 or x (x − 2) = 0
whence x = 0, 2. (See figure at bottom of column.)
• We have d V = A d x = πro2 − πri2 d x
2 d x = π 4x 2 − x 4 d x.
= π (2x)2 − x 2
0
+
2
8
3
• The volume
is
R
2
V = π 02 4x 2 − x 4 d x = π 43 x 3 − 15 x 5 0
32
5−3
3
= 64
= π 32
3 − 5 = 32π
15
15 π ≈ 13.40 cm .
10. (c) First compute an antiderivative, then apply the FTC.
− 2 = 2 cm2 .
dv = f 00 (x) d x
. Then
v = f 0 (x)
R
R
x f 00 (x) d x = x f 0 (x) − f 0 (x) d x = x f 0 (x) − f (x).
0
=
=
=
=
sin2 x cos2 x d x
R5
5
• Hence 1 x f 00 (x) d x = x f 0 (x) − f (x) 1
= 5 f 0 (5) − f (5) − f 0 (1) − f (1)
= 5 (2) − 3 − (−1 − 4) = 7 + 5 = 12.
1
1
0
2 (1 − cos 2x) · 2 (1 + cos 2x) d x
1 R π/2
1 − cos2 2x d x
4 0
R
1 π/2 1 − 1
4 0
2 (1 + cos 4x) d x
1 1 R π/2
1 − cos 4x d x
4 · 2 0
1 x − 1 sin 4x π/2 = π − 0 = π .
8
4
16
16
0
X1A/3
X1A/9
4
3
3
2
y
=
R π/2
u=x
du = d x
• Let
4. (e) Use half-angle identities.
R π/2
1
5. (c) We have
d V = A d x = πr 2 d x = π y 2 d x = π (e x )2 d x = πe2x d x.
Therefore
3
R
V = 03 πe2x d x = 12 πe2x 0 = π2 e6 − 1 ≈ 621.13 cm3 .
0
dv = x d x
.
v = 12 x 2
1
x
2
x ln x d x = 12 x 2 ln x −
R
1x
2
X1A/5
20
thickness dx
15
d x = 12 x 2 ln x − 14 x 2 .
10
5
3
• Hence 1 x ln x d x = 12 x 2 ln x − 14 x 2 1
= 92 ln 3 − 94 − − 14 = 92 ln 3 − 2.
R3
0
25
Then
0
0
1
2
x
1
thickness dx
0
0
y
R
u = ln x
du = 1x d x
2
1
6. (a) First compute an antiderivative, then apply the FTC.
• Let
cos3 x + C
9. (d) Find where the curves intersect, draw a sketch, then set
up and compute the integral for the volume.
R2 2
x − 1 − 0 d x, computed thus:
1
1
3
y
Z
cos5 x −
8. (e) We have
2
R2
R2 x
1
1 x+1 d x = 1 1 − x+1 d x = (x − ln |x + 1|) 1
= (2 − ln 3) − (1 − ln 2) = 1 + ln 2 − ln 3 = 1 + ln 23 .
2. (d) Via Hooke’s Law we have F(x) = kx or 3 = k 16 ,
2
R
whence k = 18. Thus W = 02 18x d x = 9x 2 0 = 36 ft-lb.
3. (a) The area is A =
1
5
3
1
x
2
11. A diagram of the tank, its semicircular end, and a rectangular
differential layer of water appears at bottom right.
15. We’ll integrate the rational function via partial fractions.
• Split the integrand into a sum of partial fractions.
• Clearly the layer is 3 m long, but how wide is it? Look
at a diagram of the semicircular
tank. The
p end of thep
2
2
width of the layer is 2y = 2 1 − z = 2 1 − z 2 .
• The rectangular layer of water has an area of
p
p
A = LW = 3 2 1 − z 2 = 6 1 − z 2 .
−1
d V = A d y = s 2 d y = (2x)2 d y = 4x 2 d y = 4(1 − 14 y 2 ) d y.
R
R2
4 − y 2 d y = 2 02 4 − y 2 d y
Hence the total volume is −2
2
3
= 2 4y − 13 y 3 0 = 2 8 − 83 − 0 = 32
3 ≈ 10.67 cm .
=
(A + B) x 2 + (C − B) x + (9A − C)
x
3 +C
X1A/11: Semicircular back end of tank
0
1
z
z
X1A/11: Tank
−0.5
y
−1
13. Find where the curves intersect, draw a sketch, then set up
and compute the integral for the volume.
√
• Now x = 2 − x implies x = 4 − 4x + x 2 or
x 2 − 5x + 4 = (x − 1) (x − 4) = 0, whence x = 1, 4.
Toss out x = 4. See the figure in the next column.
R
• Via cylindrical shells, we have V = 2πr h d y.
R
V = 01 2π y (2 − y) − (y 2 ) d y
R
= 2π 01 2y − y 2 − y 3 d y
1
= 2π y 2 − 13 y 3 − 14 y 4 0 .
5π
≈ 2.62 cm3 .
= 2π 1 − 13 − 14 − 0 =
6
z
y
−1 −0.5
x
X1A/12
2
0.5
1
X1A/13
thickness dy
1
0
y
y
0
y
thickness dy
1
−1
0
−2
−1
0
x
1
0
X1A/14
14. Let x = 5 sin θ . Then d x = 5 cos θ dθ . Hence (see figure at
bottom of the next column)
R 5 cos θ dθ
R
√d x
=
25 sin2 θ ·5 cos θ
x 2 25−x 2
1 R
= 25 csc2 θ dθ
=
=
0x 2 + 0x + 10
• Now integrate term-by-term.
Z
10
dx
(x − 1) x 2 + 9
Z
x
1
1
− 2
− 2
dx
=
x −1 x +9 x +9
= ln |x − 1| − 12 ln x 2 + 9 − 13 tan−1
12. The volume of a differential cross-section is
5
x
θ
1 cot θ + C
− 25
√
2
− 25−x
25x
10
A
Bx + C
+ 2
x −1
x +9
A x 2 + 9 + (Bx + C) (x − 1)
10
x
1
1
=
−
−
.
x − 1 x2 + 9 x2 + 9
(x − 1) x 2 + 9
• The work required to pump the water out of the tank is
R
R0
1/2
W =
d W = −1 −9800 (6) z 1 − z 2
dz
3/2 0
= (9800) (6) 12 23 1 − z 2
=
=
• Thus A + B = 0, C − B = 0, and 9A − C = 10.
Adding the first two equations we obtain A + C = 0,
whence C = −A. Substituting this into the third
equation gives 10 A = 10 whence A = 1, C = −1,
and B = C = −1. Therefore,
Its thickness is dz. Here are the volume of the layer,
its weight, and the work required to lift it to the top of
the tank. Recall that δ = ρg = 9800 N/m3 .
p
d V = A dz = 6 1 − z 2 dz
p
d F = δ d V = 9800 (6) 1 − z 2 dz
p
d W = d F (0 − z) = −9800 (6) z 1 − z 2 dz
= (9800) (2) − 0 = 19,600 J.
10
(x − 1) x 2 + 9
2 1/2
(25 − x )
+C
2
1
x
2