Math 142, Midterm 1 a Solutions
October 24, 2011
Part I. (20 minutes, 20 points) There are 2 problems, do both of them. No calculator.
1. Circle T for true or F for false, or fill in the blank line. (Work on scratch paper.)
a) The area of the region in the first quadrant of the xy-plane that is bounded by the two
curves y = x and y = sin(x) and the vertical line x = π/2 is π 2 /4 − 2 .
π/2
R π/2
= π 2 /8 − 1
A = 0 x − sin(x) dx = 12 x2 + cos(x)
T
F
0
b) The average value of x cos(x2 ) over the interval [0, π ] is
sin(π 2 )/2π
(exact value please).
π
Rπ
Ave Value = π1 0 x cos(x2 ) dx = π1 · 12 sin(x2 ) = sin(π 2 )/2π
0
R √
c) x 1 − 3x2 dx can be evaluated easily using the method of
substitution: u = 1 − 3x2
.
R
d) To evaluate tan2 (x) sec4 (x) dx split off sec2 (x) and then
convert the remaining sec2 (x) to tangents
then substitute u = tan(x)
e) If the planar region described in part a is rotated about the vertical line x = −1, then its
R π/2
volume is given by the integral 0 2π(x + 1)(x − sin(x)) dx.
T
.
F
2. Do part a or part b, not both. Show all of your work in the space below and box your answer. Neatness counts.
a) A 10 foot chain that weighs 4 pounds extends from the top of a 10 foot table to the floor where it is
attached to an 8 pound weight. Find the work required to pull the chain and the weight up to the top of
the table.
Let the y-axis extend from the floor, y = 0, to the table top, y = 10. When the weight is at the position y,
the force to move it upward is F (y) = δ(10 − y) + 8 pounds, where δ = 4/10 pounds per foot is the linear
density of the chain. Therefore, the work required to pull the weight from the floor to the table top is
Z 10
Z 10
Z 10
10
4
W =
F (y) dy =
(10
−
y)
+
8
dy
=
12 − 52 y dy = 12y − 15 y 2 = 120 − 20 = 100 ft-lbs .
10
0
0
0
0
b) If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, then how much work is
needed to stretch it 9 in. beyond its natural length?
1
R1
Find the spring constant using the equation 12 = 0 kx dx = 21 kx2 0 = 12 k, so k = 24 lbs per foot.
The work required to stretch the spring 9 inches (= 3/4 feet) is
Z
W =
3/4
Z
kx dx =
0
0
3/4
3/4
24x dx = 12x2 = 12 ·
0
9
16
=
27
4
ft-lbs .
Part II. (30 minutes, 30 points) There are 4 problems, do the first one and 2 more for a total of 3.
1. (You must do this problem.) Evaluate the following integrals, derivatives, or limits, showing all the steps in
your calculation. Neatness counts. No calculator.
Z
a)
Z
sin(x)
dx
1 + cos2 (x)
b)
ln(x)
dx
x2
Integrate by parts: u = ln(x) , dv = x−2 dx, so
du = 1/x , v = −x−1 , and
Z
Z
ln(x)
dx
=
−
ln(x)/x
+
x−2 dx .
I=
x2
Substitute u = cos(x), du = − sin(x) dx to obtain
Z
Z
sin(x)
−du
I=
.
dx
=
2
1 + cos (x)
1 + u2
Therefore,
Therefore,
I = − arctan(u) + C = − arctan(cos(x)) + C .
I = − ln(x)/x − 1/x + C .
c)
Z
p
d ln t 1 − t2
dt
d)
0
Simplify the log term first:
p
ln t 1 − t2 = ln(t) +
1
2
ln(1 − t2 ) .
1
t
1 − 2t2
1 1 −2t
+ ·
=
−
=
.
t
2 1 − t2
t
1 − t2
t(1 − t2 )
lim x ln(x2 )
Move the x term into the denominator as x−1 to
make it an ∞/∞ indeterminant form:
lim+ x ln(x2 ) = lim+
x→0
Therefore,
1+cos(1)
I = − ln(u)2
= ln(2) − ln(1 + cos(1)) .
f)
x→0+
x→0
2 ln(x)
2/x
= lim+
.
x−1
x→0 −x−2
Therefore,
sin(x)
dx
1 + cos(x)
Substitute u = 1 + cos(x), du = − sin(x) dx to
obtain
Z 1
Z 1+cos(1)
sin(x)
−du
I=
dx =
.
u
0 1 + cos(x)
2
Therefore, the derivative is
e)
1
lim x1/x
x→∞
Convert to base e:
1
lim x1/x = lim e x ln(x) .
x→∞
Now apply L’Hospital’s Rule to the term in the
exponent (it is indeterminant of type ∞/∞).
lim x ln(x2 ) = −2 lim+ x = 0 .
x→0+
x→∞
lim
x→∞
x→0
ln(x)
1/x
= lim
= 0.
x→∞ 1
x
Therefore,
lim x1/x = e0 = 1 .
x→∞
2. Cobalt 60 has a half-life of 5.24 years. Make the following calculations, showing all of your work.
a) Find the mass that remains from a 100 mg sample after 20 years.
The model for half-life exponential decay can be expressed as
M (t) = 100
t/5.24
1
.
2
Therefore,
20/5.24
1
M (20) = 100
= 7.0963 mg .
2
b) How long would it take for the 100 mg mass to decay to 1 mg?
1 t/5.24
2
Solve the equation M (t) = 1: 100
Therefore,
t = 5.24 ·
= 1 =⇒
1 t/5.24
2
=
1
100
=⇒
t
5.24
ln(1/2) = ln(1/100) .
ln(100)
= 34.814 years .
ln(2)
3. The function f (x) = x ln(x) is strictly increasing on the interval [e−1 , ∞) (no proof required). Let g denote its
inverse function.
[−e−1 , ∞)
a) The domain of g is
b) f (e) =
e
, f (1) =
g(e) =
e
.
. The range of g is
0
[e−1 , ∞)
.
, and
Approximate g(1) using the graph of f on the right.
g(1) ≈
1.75
.
A small portion of the graph of f (x) = x ln(x).
c) Find g 0 (e) and approximate g 0 (1).
Note that f 0 (x) = 1 + ln(x). Therefore,
g 0 (e) =
1
1
=
f 0 (e)
2
and g 0 (1) ≈
1
1
=
= 0.638 .
f 0 (1.75)
1 + ln(1.75)
4. Let R denote the region between the graph of the
function f (x) = sin3 (x) and the x-axis from x = 0 to
x = π. See the picture.
Obtain exact answers to a, b, and c below. Show each
step in each calculation.
Neatness counts. No calculators.
The region R.
a) Find the area of R.
Z
Area =
π
sin3 (x) dx =
0
Z
π
sin2 (x) sin(x) dx
0
Z
π
(1 − cos2 (x)) sin(x) dx
π
= − cos(x) + 31 cos3 (x)
0
= 1 − 31 − −1 + 13
=
0
=2−
2
3
=
4
3
.
b) Use your answer to part a to find the average value of sin3 (x) over the interval [0, π].
Average Value =
1 4
4
· =
.
π 3
3π
c) Find the volume of the solid obtained when the region R is rotated about the y-axis.
Rπ
Helpful hint. By symmetry, 0 cos3 (x) dx = 0.
The volume formula can be obtained using cylindrical shells.
Z π
Volume =
2πx sin3 (x) dx
0
3
Applying integration by parts, u = x, dv = sin (x) dx, so du = dx and, using the antiderivative of sin3 (x)
in part a, v = 31 cos3 (x) − cos(x) . Consequently,
Z π
Volume =
2πx sin3 (x) dx
0
Z π
π
3
3
1
1
= 2π x 3 cos (x) − cos(x) −
3 cos (x) − cos(x) dx
0
0
Z π
= 2π π − 13 + 1 +
cos(x) dx
0
= 2π 2 ·
Note that
Rπ
0
2
3
4π 2
=
.
3
cos(x) dx is also 0 by the symmetry of its graph.