Solutions to Test # 2
1. Use the Comparison
Z ∞ Theorem to determine if the following improper integral is con1
vergent or divergent:
dx
x + ex
0
Solution. We have:
x + ex > ex
1
1
06
6
x + ex
ex
It is easy to see that
Z
for x > 0
∞
−x t
1
dx
=
lim
−e 0 = 1
t→∞
ex
0
Z ∞
1
is convergent, so by the Comparison Theorem
dx is also convergent.
x
0Z x + e
π/2
cos x
√
dx
2. Evaluate the following improper integral:
sin x
0
Z
cos x
√
Solution. Let us first evaluate the following indefinite integral:
dx. Using the
sin x
substitution u = sin x, du = cos x dx, we get:
Z
Z
√
√
cos x
1
√
√ du = 2 u + C = 2 sin x + C.
dx =
u
sin x
Therefore:
Z
0
π/2
cos x
√
dx = lim+
t→0
sin x
Z
t
π/2
h √
iπ/2
cos x
√
dx = lim+ 2 sin x
=2
t→0
t
sin x
3. If f (x) = sin x, find the average value of f on [0, π].
Solution.
Z π
1
1
2
fave =
sin x dx = [− cos x]π0 = .
π−0 0
π
π
4. (a) Sketch the region enclosed by the curves y = x4 − x2 and y = 1 − x2 . (b) Find the
area of the region from part (a).
Solution. (a).
1
0.8
0.6
0.4
0.2
-1
-0.5
0.5
-0.2
1
1
(b). First we need to find the intersection points:
x4 − x2 = 1 − x2
x4 = 1
x = ±1
Now we can use top-bottom formula:
Z 1
Z
2
4
2
A=
(1 − x ) − (x − x ) dx =
1
8
x5
= .
1 − x dx = x −
5 −1 5
−1
−1
1
4
5. (a) Sketch the region bounded by the curves y = 6 − x2 and y = 1. (b) Find the
volume of the solid obtained by rotating the region from part (a) about the x-axis.
Solution. (a)
6
4
2
-3
-1
-2
1
2
3
-2
When the above region is rotated about the x-axis, the resulting solid looks like a donut:
2 1 0
-1 -2
5
0
-5
-5
-2.5
0
2.5
5
2
(b) First let us find the area of a cross-section (a cross section will be a ring with inner
radius equal to 2 and outer radius equal to 6 − x2 ):
A(x) = π(6 − x2 )2 − π(2)2 = π(36 − 12x2 + x4 − 4) = π(32 − 12x2 + x4 )
Now we can find points at which y = 6 − x2 and y = 2 intersect:
6 − x2 = 2
x2 = 4
x = ±2
So the volume is:
2
x5
384
3
V =
π(32 − 12x + x ) dx = π 32x − 4x +
=
π.
5 −2
5
−2
Z
2
2
4
6. Find the length of the curve given by the following system of parametric equations:
x = t3 − 3t, y = 3t2 , 0 6 t 6 1.
Solution. We have:
dx
dy
= 3t2 − 3
= 6t
dt
dt
2 2
dy
dx
+
= 9t4 − 18t2 + 9 + 36t2 = 9(t4 + 2t2 + 1) = 9(t2 + 1)2
dt
dt
Therefore:
Z 1p
Z 1
1
2
2
L=
9(t + 1) dt =
3(t2 + 1) dt = t3 + 3t 0 = 4.
0
0
7. A heavy rope, 20 meters long, weighs 1 kg per meter and hangs over the edge of
a building 30 meters high. How much work is done in pulling the rope to the top of the
building? (g ≈ 9.81 sm2 ).
Solution. Suppose x meters have already been pulled to the top. Then the remaining
20 − x meters (whose mass is m(x) = (20 − x) · 1 kg) generates a force F (x) = m(x)g =
(20 − x)g N. Therefore the work needed to pull the rope is equal to
Z 20
Z 20
W =
F (x) dx =
(20 − x)g dx = 200g ≈ 1962 J.
0
0
8. A force of 30 N is required to maintain a spring stretched from its natural length of
12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to
20 cm?
Solution. We know that force required to hold the spring stretched x cm beyond its
natural length is given by F (x) = kx. Therefore
k · (15 − 12) = 30
k = 10 N/cm
So F (x) = 10x. The work needed to stretch the spring from 12 cm to 20 cm is given by
Z 20−12
8
W =
10x dx = 5x2 0 = 320 N·cm = 3.2 J.
0
3