MATH 172.504
Examination 1 Solutions
February 16, 2012
Z
1.
2
6x
Evaluate
p
2x2 + 1 dx.
0
(A) 26
(B) 27
(C) 28
(D) 29
(E) 30
Solution: A We make the substitution u = 2x2 + 1, du = 4x dx, and find (remembering
the substitute the limits as well)
Z
2
Z
p
2
6x 2x + 1 dx =
0
2.
1
9
6√
3 h 2 3/2 i9
u du =
u
= 27 − 1 = 26.
4
2 3
1
Consider the region in the first quadrant bounded by y = x2 , y = 9, and x = 0. Which
integral below gives the volume of the solid obtained by rotating this region about the
x-axis?
Z 9
(A) π
(9 − x2 ) dx
0
Z
3
x(9 − x2 ) dx
(B) 2π
0
Z
(C) 2π
Z
(x4 − 9) dx
0
3
(D) π
Z
9
(81 − x4 ) dx
0
3
(E) π
y 2 dy
0
Solution: D We use washers, noting that the horizontal line y = 9 sits above the parabola
y = x2 on the interval 0 ≤ x ≤ 3. So the outer radius at a point x is 9, and the inner radius
is x2 .
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
= 25(0.2)2 = 1.
50x dx = 25x2
W =
0
0
Z
4.
Let F (x) =
xp
t3 + 1 dt. Which of the following statements about F (x) are true?
1
I. F (1) = 0.
II. F 0 (2) = 2.
III. F 0 (2) = 3.
(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 the Fundamental Theorem of Calculus says that
√
√
F 0 (x) = x3 + 1, and so F 0 (2) = 9 = 3. Thus III is true but II is false .
5.
What is the average value of f (x) = cos(3x) on the interval 0 ≤ x ≤ π2 ?
2
(A) −
3π
3
(B) −
2π
1
(C)
3π
1
(D)
2
(E) 0
Solution: A We need to calculate the integral
π/2
Z
2 π/2
2 1
2
3π
2
cos(3x) dx =
sin(3x)
=
sin
=− .
π 0
π 3
3π
2
3π
0
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
4
Z 4
y2
y3
16
56
y2
A=
(8 − y) −
dy = 8y −
−
= 32 − 8 −
=
.
4
2
12 0
3
3
0
2
7.
Evaluate the following integrals.
Z π/3
4 cos2 θ sin θ dθ
(a)
0
Z
√
(b) x2 x + 1 dx
e2x
dx
e2x + 1
Solution: (a) We substitute u = cos θ, du = − sin θ dθ, giving
Z
(c)
Z
π/3
1/2
Z
2
4 cos θ sin θ dθ = −
0
1
4 3
4u du =
u
3
2
1
1/2
4 4
= −
3 3
1
7
=
.
8
6
(b) We substitute u = x + 1, du = dx, x = u − 1, and find
Z
Z
Z
√
√
2√
2
x x + 1 dx = (u − 1) u du = (u2 − 2u + 1) u du
Z
= (u5/2 − 2u3/2 + u1/2 ) du
2
4
2
= u7/2 − u5/2 + u2/3 + C
7
5
3
2
4
2
= (x + 1)7/2 − (x + 1)5/2 + (x + 1)3/2 + C .
7
5
3
(c) We substitute u = e2x + 1, du = 2e2x dx, and find
Z
Z
1
1
1
e2x
dx =
du = ln |u| + C =
2x
e +1
2
u
2
8.
1
2
ln(e2x + 1) + C .
Let R be the region in the first quadrant of the xy-plane bounded by the curves
πx y = cos
, y = 2 − 2x2 , and the y-axis.
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 y-axis.
(c) The line y = −1.
Solution: In the first quadrant the parabola sits above the cosine curve over the interval
[0, 1].
(a) Here we use washers:
Z
V =π
1h
(2 − 2x2 )2 − cos2
0
3
πx i
2
dx
(b) Here we use cylindrical shells:
Z
V = 2π
0
1
h
πx i
x (2 − 2x2 ) − cos
dx
2
(c) Here we use washers again:
Z
V =π
0
9.
1
πx 2 (2 − 2x2 + 1)2 − cos
dx
+1
2
[8 points] Let S be the 3-dimensional solid whose base is the region in the first quadrant
of the xy-plane bounded by y = cos x, the x-axis, and the y-axis, and whose cross-sections
perpendicular to the x-axis are squares. Write down an integral for the volume of S. (You
do not need to evaluate the integral.)
Solution: The region sits below y = cos x over the interval 0 ≤ x ≤ π2 . At a point x in the
interval, the cross-section of S is then a square of side-length cos x. Therefore,
Z
V =
0
4
π/2
cos2 x dx .