MATH 172.200
Examination 1 Solutions
February 16, 2012
Z
1.
π/3
4 cos2 θ sin θ dθ.
Evaluate
0
(A) 0
(B) 1
1
(C)
3
7
(D)
6
1
6
Solution: D We first substitute u = cos θ, du = − sin θ dθ, giving
(E) −
Z
π/3
4 cos2 θ sin θ dθ = −
Z
0
2.
What is the average value of f (x) =
(A)
(B)
(C)
(D)
(E)
1/2
4u2 du =
1
x2
4 3
u
3
1
=
1/2
4 4
−
3 3
1
7
= .
8
6
x
on the interval 0 ≤ x ≤ 2?
+4
1
ln(2)
4
1
ln(2)
2
1
ln(8)
4
1
ln(8)
2
1
ln(4)
2
Z
1 2 x
Solution: A We need to calculate the definite integral
dx, so we substitute
2 0 x2 + 4
u = x2 + 4, du = 2x dx:
Z
Z
i8 1
1 81
1 2 x
1h
1
1
dx
=
du
=
ln
|u|
= ln(8) − ln(4) = ln(2).
2 0 x2 + 4
4 4 u
4
4
4
4
4
3.
Our ideal spring follows Hooke’s Law. It requires 5 Newtons of force to be held 10 cm from
rest. How much work (in Joules) is done in stretching the spring from rest to 20 cm?
(A) 0
(B) 1
(C) 10
(D) 100
(E) 1000
1
Solution: B We first need to calculate the spring constant. We are given that the force
at 0.1 meters (= 10 cm) is F (0.1) = 5. Hooke’s Law asserts that F (x) = kx for a constant
k, so k(0.1) = 5, which implies that k = 50. Therefore to calculate the work we evaluate
Z 0.2
h
i0.2
W =
50x dx = 25x2
= 25(0.2)2 = 1.
0
0
Z
4.
x2
Let F (x) =
p
t3 + 17 dt. Which of the following statements about F (x) are true?
1
I. F (1) = 0.
II. F 0 (2) = 9.
III. F 0 (2) = 36.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Solution: E Since F (1) is equal to a definite integral over an interval of length 0, its value
must be zero. Therefore, I is true . Now if we let
Z xp
G(x) =
t3 + 17 dt,
1
√
then the Fundamental Theorem of Calculus says that G0 (x) = x3 + 17. Since F (x) =
G(x2 ), it follows from the chain rule that F 0 (x) = G0 (x2 ) · (2x). Therefore,
√
F 0 (2) = G0 (4) · 4 = ( 81) · 4 = 36.
So III is true but II is false .
5.
Evaluate the limit
6 4
16 36
4(n − 1)2
lim
+ 2 + 2 + ··· +
+4 .
n→∞ n n2
n
n
n2
(A) 0
2
(B)
3
8
(C)
3
(D) 4
(E) 8
Solution: E We strongly suspect that this is the limit of Riemann sums, so we look for
the definite integral the limit represents. What we see is that, for 1 ≤ i ≤ n, the i-th term
in the sum is
2
4 · i2
2i
=
.
n2
n
2
Thus function most likely is related to x2 . The interval we are integrating over is of length
2, ending at 2, so it must be [0, 2]. Thus we suspect the integral could be related to
Z
2
x2 dx.
0
However, upon closer inspection the n6 in front of the limit has an extra factor of 3, so that
2
6
n = 3 · ∆x = 3 · n , and so the integral must be
2
Z
h i2
3x2 dx = x3 = 8.
0
0
Converting this integral into the limit of Riemann sums using right-hand end points confirms
this to be the case.
6.
Let R be the region in the first quadrant of the xy-plane bounded above by the union of
√
the curve y = 2 x and the line y = −x + 8 and bounded below by the x-axis. Find the
area of R by setting up the integral with respect to y.
√
Solution: When we consider the region in question, we have the parabola y = 2 x to the
left of the line y = −x + 8. The two curves intersect in the first quadrant at (4, 4). Now
converting to y, the parabola is x = y 2 /4 and the line is x = 8 − y, and so we find the area
by integrating
Z
A=
0
7.
4
y2
(8 − y) −
4
y2
y3
dy = 8y −
−
2
12
4
= 32 − 8 −
0
16
56
=
.
3
3
Evaluate the following indefinite integrals.
Z
(a) tan(2x) ln(cos(2x)) dx
Z
(b)
x2
dx
x6 + 1
e6x
dx
e2x + 1
Solution: (a) First we substitute u = cos(2x), du = −2 sin(2x) dx, and so
Z
Z
1
ln u
tan(2x) ln(cos(2x)) dx = −
du.
2
u
Z
(c)
We substitute again, taking v = ln u, dv =
−
1
2
Z
ln u
1
du = −
u
2
Z
1
u
du, obtaining
2
1
1
v dv = − v 2 + C = − ln(cos(2x)) + C .
4
4
(b) We substitute u = x3 , du = 3x2 dx, and find
Z
x2
1
dx =
x6 + 1
3
Z
1
1
1
du = tan−1 (u) + C = tan−1 (x3 ) + C .
u2 + 1
3
3
3
(c) We substitute u = e2x + 1, du = 2e2x dx, and start by obtaining
Z
Z 4x
e6x
e
1
dx
=
du.
2x
e +1
2
u
Since e4x = (e2x )2 = (u − 1)2 , we find
Z
Z
Z 2
e6x
(u − 1)2
u − 2u + 1
1
1
dx =
du =
du
2x
e +1
2
u
2
u
Z 1
1
u−2+
du
=
2
u
1 u2
=
− 2u + ln |u| + C
2 2
=
8.
1
1 2x
(e + 1)2 − (e2x + 1) + ln(e2x + 1) + C .
4
2
Let R be the region in the first quadrant of the xy-plane bounded by the curves
πx , y = 2 − 2x2 , and the y-axis.
y = cos
2
Set up integrals with respect to x (do not evaluate!) that express the volumes of the solids
obtained by rotating R about the following axes of rotation. (It may help to start by
sketching the region R.)
(a) The x-axis.
(b) The line y = −2.
(c) The line x = 2.
Solution: In the first quadrant the parabola sits above the cosine curve over the interval
[0, 1].
(a) Here we use washers:
1h
Z
V =π
(2 − 2x2 )2 − cos2
0
πx i
2
dx
(b) Here we use washers again:
Z
1
2
2
(2 − 2x + 2) − cos
V =π
0
πx 2
+2
2 dx
(c) Here we use cylindrical shells:
Z
V = 2π
0
1
h
πx i
(2 − x) (2 − 2x2 ) − cos
dx
2
4
9.
A tank the shape of a right circular cone has base diameter 6 meters and height 4 meters.
The tank also has a spout of height 2 meters that extends from the apex of the cone. If
the tank is full of water of density 1000 kg/m3 , set up an integral to express the amount
of work required (in Joules) to lift the water out of the tank to the level of the top of the
spout. You may assume that the acceleration of gravity is 9.8 m/s2 . (You do not need to
evaluate the integral.)
Solution: We let y = 0 be the bottom of the cone and y = 4 be the top and consider a
circular layer of water at level y in the tank. Using similar right triangles, we see that the
radius of this layer is related to y by
4−y
4
= ,
r
3
and so r = 3 − 34 y. Therefore the volume of the infinitesimal layer is
3 2
dV = πr2 dy = π 3 − y
dy.
4
The distance the layer travels to get to the top of the spout is 6 − y, and so taking into
account the density of the water and the acceleration of gravity we find that
3 2
9800π 3 − y (6 − y) dy .
4
4
Z
W =
0
It is possible to change our orientation at the beginning, and let y = 0 be the top of the
cone and y = 4 be the bottom. If we had done this, the resulting integral would have been
Z
W =
4
9800π
0
3y
4
2
Ultimately the two integrals have the same value.
5
(2 + y) dy .