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Math 211 7–27–
Review Problems for Test 2
These problems are provided to help you study. The presence of a problem on this sheet does not imply that a similar problem will appear on the test. And the absence of a problem from this sheet does not imply that the test will not have a similar problem.
- Find the area of the region bounded by the graphs of y = x^2 − 3 x and y = 15 − x.
- Find the area of the region between y = x^2 − x and y = x + 8 from x = 0 to x = 5.
- The region bounded by y = 4x − x^2 and the x-axis is revolved about the x-axis. Find the volume of the solid that is generated.
- Consider the region in the x-y plane bounded by y = ex, the line y = 1, and the line x = 1. Find the volume generated by revolving the region:
(a) About the line y = 1.
(b) About the line x = 2.
(c) About the line y = e.
- The base of a solid is the region in the x-y plane bounded by the curves y = x^2 and y = x + 2. The cross-sections of the solid perpendicular to the x-y plane and the x-axis are isosceles right triangles with one leg in the x-y plane. Find the volume of the solid.
- The region which lies above the x-axis and below the graph of y =
x^2 + 1
, −∞ < x < ∞, is revolved
about the x-axis. Find the volume of the solid which is generated.
Hint: (^) ∫ 1 (x^2 + 1)^2
dx =
x x^2 + 1
tan−^1 x + C.
- The base of a rectangular tank is 2 feet long and 3 feet wide; the tank is 6 feet high. Find the work done in pumping all the water out of the top of the tank.
- Write a formula for the n-th term of the sequence, assuming that the terms continue in the “obvious” way.
(a) 7, 11 , 15 , 19 , 23 , 27 ,.. ..
(b)
- Determine whether the sequence converges or diverges; if it converges, find the limit.
(a) { 1. 0001 n}.
(b)
en^ + 3n 2 n^ + πn
(c)
2 n^3 − 5 n + 7 7 n^2 − 13 n^3
(d)
(arctan n)^2
- A sequence is defined recursively by
a 1 = 5, an+1 =
6 an + 27 for n ≥ 1.
Find lim n→∞ an.
- In each case, determine whether the series converges or diverges. You should cite the test you’re using by name (to avoid ambiguity); be sure you verify that the hypotheses of the test apply.
(a)
∑^ ∞
k=
3 k^ + 2k 6 k^
(b)
∑^ ∞
k=
k(ln k)^2
(c)
∑^ ∞
k=
k^2 − 3 k + 2 k^4
(d)
∑^ ∞
k=
k^2 − 1
(e)
∑^ ∞
k=
3 k + 5
(f)
∑^ ∞
k=
k + 1 k^2 + 2
(g)
∑^ ∞
k=
sin
k
(h)
∑^ ∞
k=
2 k 3 k^ + 2
(i)
∑^ ∞
i=
3 i i^3
(j)
∑^ ∞
k=
2 k + 2 2 k − 1
)k^2 .
(k)
∑^ ∞
k=
arctan k.
(l)
∑^ ∞
k=
(k!)^2 (2k + 1)!
- (a) Find the partial fractions decomposition of
(2k + 1)(2k + 3)
(b) Use (a) to find the sum of the series
∑^ ∞
k=
(2k + 1)(2k + 3)
Since the curves cross between 0 and 5, I will need two integrals. On the left-hand piece, the top curve is y = x + 8 and the bottom curve is y = x^2 − x. On the right-hand piece, the top curve is y = x^2 − x and the bottom curve is y = x + 8. The area is ∫ (^4)
0
(x + 8) − (x^2 − x)
dx +
4
(x^2 − x) − (x + 8)
dx =
0
(−x^2 + 2x + 8) dx +
4
(x^2 − 2 x − 8) dx =
[ −
x^3 + x^2 + 8x
] 4
0
[
x^3 − x^2 − 8 x
] 5
4
- The region bounded by y = 4x − x^2 and the x-axis is revolved about the x-axis. Find the volume of the solid that is generated.
1 2 3 4
1
2
3
The region extends from x = 0 to x = 4. I’ll use circular slices. The radius of a typical slice is r = y = 4x − x^2. The area of a typical slice is
πr^2 = π(4x − x^2 )^2 = π(16x^2 − 8 x^3 + x^4 ).
The volume generated is
V =
0
π(16x^2 − 8 x^3 + x^4 ) dx = π
[
x^3 − 2 x^4 +
x^5
] 4
0
512 π 15
- Consider the region in the x-y plane bounded by y = ex, the line y = 1, and the line x = 1. Find the volume generated by revolving the region:
(a) About the line y = 1.
dx
r = e - 1x
Since the solid has no “holes” or “gaps” in its interior, I can use circular slices. The radius of a slice is r = ex^ − 1, so the volume is
V =
0
π(ex^ − 1)^2 dx = π
0
(e^2 x^ − 2 ex^ + 1) dx = π
[
e^2 x^ − 2 ex^ + x
] 1
0
πe^2 2
− 2 πe +
5 π 2
(b) About the line x = 2.
dx
1
h = e - 1x
x = 2
x r = 2 - x
2 I’ll use cylindrical shells. The height is h = ex^ − 1, and the radius is r = 2 − x. The volume is
V =
0
2 π(ex^ − 1)(2 − x) dx = 2π
0
(2ex^ − 2 − xex^ + x) dx = 2π
[
2 ex^ − 2 x − xex^ + ex^ +
x^2
] 1
0
4 πe − 9 π ≈ 5. 88460. Here’s the work for part of the integral:
d dx
dx
- x ex ց − 1 ex ց
- 0 → ex ∫ xex^ dx = xex^ − ex^ + C.
(c) About the line y = e.
dy
1
1
x h
h = 1 - x = 1 - ln y
r = e - y
y
e
I’ll use cylindrical shells. Since y = ex^ gives x = ln y, the height is h = 1 − x = 1 − ln y, and the radius is r = e − y. The vertical limits on the region are y = 1 and y = e. The volume is
V =
∫ (^) e
1
2 π(1 − ln y)(e − y) dy = 2π
∫ (^) e
1
(e − e ln y − y + y ln y) dy =
2 π
[
ey − ey ln y + ey −
y^2 +
y^2 ln y −
y^2
]e
1
3 πe^2 2
− 4 πe +
3 π 2
- The region which lies above the x-axis and below the graph of y =
x^2 + 1
, −∞ < x < ∞, is revolved
about the x-axis. Find the volume of the solid which is generated.
0
1 -
-0.
0
1
Chop the solid up into circular slices perpendicular to the x-axis. The thickness of a typical slice is dx.
The radius of a slice is r =
x^2 + 1
. The volume is
V =
−∞
π ·
(x^2 + 1)^2
dx =
0
π ·
(x^2 + 1)^2
dx +
−∞
π ·
(x^2 + 1)^2
dx.
Compute the first integral: ∫ (^) ∞
0
π ·
(x^2 + 1)^2
dx = lim a→+∞
∫ (^) a
0
π ·
(x^2 + 1)^2
dx = π · lim a→+∞
[
x x^2 + 1
tan−^1 x
]a
0
π 2
lim a→+∞
a a^2 + 1
π^2 4
(I used the fact that lima→+∞ tan−^1 a =
π 2
Similarly, (^) ∫ 0
−∞
π ·
(x^2 + 1)^2
dx =
π^2 4
The volume is
π^2 4
π^2 4
π^2 2
- The base of a rectangular tank is 2 feet long and 3 feet wide; the tank is 6 feet high. Find the work done in pumping all the water out of the top of the tank.
Divider the water up into rectangular slabs parallel to the base. Let y denote the height of a slab above the base.
dy
y 2
3
6
6 - y
The volume of a typical slab is (2)(3) dy = 6 dy, so the weight is 62. 4 · 6 dy. (The density of water is 62 .4 pounds per cubic foot.) A slab at height y must be lifted a distance of 6 − y to get to the top of the tank. Therefore, the work done in lifting the slab is 62. 4 · 6(6 − y) dy. The total work is
∫ (^6)
0
- 4 · 6(6 − y) dy = 62. 4 · 6
[
6 y −
y^2
] 6
0
= 6739.2 foot-pounds.
- Write a formula for the n-th term of the sequence, assuming that the terms continue in the “obvious” way.
(a) 7, 11 , 15 , 19 , 23 , 27 ,.. .. an = 7 + 4n for n = 0, 1 , 2 ,....
(b)
an =
2 n 3 + 5n
for n = 1, 2 , 3 ,....
- Determine whether the sequence converges or diverges; if it converges, find the limit.
(a) { 1. 0001 n}
Since { 1. 0001 n} is a geometric sequence with ratio r = 1. 0001 > 1,
lim n→∞
- 0001 n^ = +∞.
(b)
en^ + 3n 2 n^ + πn
Divide the top and bottom by πn^ (since πn^ is the biggest exponential in the fraction):
lim n→∞
en^ + 3n 2 n^ + πn^
= lim n→∞
en pin^
3 n πn 2 n πn^
I computed the limit using the fact that
en pin^
( (^) e π
)n , 3 n πn^
π
)n , and 2 n πn^
π
)n
are geometric sequences and their ratios are all less than 1. Therefore, they go to 0 as n → ∞.
(c)
2 n^3 − 5 n + 7 7 n^2 − 13 n^3
lim n→∞
2 n^3 − 5 n + 7 7 n^2 − 13 n^3
I did this by considering the highest powers on the top and bottom; they’re both x^3 , so I just looked at their coefficients. You could also do this by using L’Hˆopital’s rule, or by dividing the top and the bottom by x^3.
(d)
(arctan n)^2
lim n→∞ (arctan n)^2 =
lim n→∞ arctan n
( (^) π 2
π^2 4
(c)
∑^ ∞
k=
k^2 − 3 k + 2 k^4
∑^ ∞
k=
k^2 − 3 k + 2 k^4
∑^ ∞
k=
k^2
∑^ ∞
k=
k^3
∑^ ∞
k=
k^4
The series on the right are convergent p-series. Hence, the original series converges.
(d)
∑^ ∞
k=
k^2 − 1
lim k→∞
k^2 − 1 1 k^2
= lim k→∞
2 k^2 k^2 − 1
The limit is a finite positive number.
∑^ ∞
k=
k^2 converges, because it’s a p-series with p = 2 > 1. Therefore,
∑^ ∞
k=
k^2 − 1
converges by Limit Comparison.
(e)
∑^ ∞
k=
3 k + 5
Let f (x) =
3 x + 5
. Then f is positive and continuous for x ≥ 1. The derivative is
f ′(x) =
(3x + 5)^2
f ′(x) < 0 for x ≥ 1, so f decreases for x ≥ 1. The hypotheses of the Integral Test are satisfied. Compute the integral: ∫ (^) ∞
1
3 x + 5 dx = lim p→∞
[
ln | 3 x + 5|
]p
1
plim→∞ (ln^ |^3 p^ + 5| −^ ln 8) = +∞.
The limit diverges, so the integral diverges. Therefore, the series diverges, by the Integral Test.
(f)
∑^ ∞
k=
k + 1 k^2 + 2
Rewrite the series as ∑∞
k=
(k + 1)^2 (k^2 + 2)^2
For large k, (k + 1)^2 (k^2 + 2)^2
k^2 k^4
k^2
Use Limit Comparison with the series
∑^ ∞
k=
k^2
. The limiting ratio is
lim k→∞
(k + 1)^2 (k^2 + 2)^2 1 k^2
= lim k→∞
k^2 (k + 1)^2 (k^2 + 2)^2
The limit is finite and positive. Since
∑^ ∞
k=
k^2
is a convergent p-series (p = 2), the series converges by
Limit Comparison.
(g)
∑^ ∞
k=
sin
k
I’ll use Limit Comparison with
∑^ ∞
k=
k^2
. Rationale: For θ ≈ 0, sin θ ≈ θ, so
sin
k
k^2
lim k→∞
sin
k
k^2
lim k→∞
sin
k 1 k
2
=
lim m→ 0
sin m m
(I set m =
k
. As k → ∞, m → 0.)
The limit is a finite, positive number, and the series
∑^ ∞
k=
k^2 is a convergent p-series (p = 2). Therefore,
the series converges, by Limit Comparison.
(h)
∑^ ∞
k=
2 k 3 k^ + 2
2 k 3 k^ + 2
2 k 3 k^
since making the bottom of a fraction smaller makes the fraction larger.
The series
∑^ ∞
k=
2 k 3 k^
is geometric with ratio r =
, so it converges. Therefore, the series
∑^ ∞
k=
2 k 3 k^ + 2
converges by comparison.
(i)
∑^ ∞
i=
3 i i^3
Apply the Ratio Test:
lim i→∞
ai+ ai
= lim i→∞
3 i+ (i + 1)^3 3 i i^3
= lim i→∞
3 i+ 3 i^
i^3 (i + 1)^3
= lim i→∞
i i + 1
- (a) Find the partial fractions decomposition of
(2k + 1)(2k + 3)
(2k + 1)(2k + 3)
A
2 k + 1
B
2 k + 3
2 = A(2k + 3) + B(2k + 1).
Set x = −
: I get 2 = 2A, so A = 1.
Set x = −
: I get 2 = − 2 B, so B = −1. Therefore, 2 (2k + 1)(2k + 3)
2 k + 1
2 k + 3
(b) Use (a) to find the sum of the series
∑^ ∞
k=
(2k + 1)(2k + 3)
∑^ ∞
k=
(2k + 1)(2k + 3)
∑^ ∞
k=
2 k + 1
2 k + 3
The second fraction in each pair cancels with the first fraction in the next pair. The only one that isn’t
cancelled is the very first one:
. Therefore,
∑^ ∞
k=
(2k + 1)(2k + 3)
- If the series
∑^ ∞
k=
ak, converges, does the series
∑^ ∞
k=
ak converge?
If the series
∑^ ∞
k=
ak converges, then the series
∑^ ∞
k=
ak converges. They only differ in the first 16 terms,
and a finite number of terms cannot affect the convergence or divergence of an infinite series.
- Does the series
∑^ ∞
k=
(−1)k+^
2 k + 1 4 k + 3
converge?
The series alternates, but lim k→∞
2 k + 1 4 k + 3
The (−1)k+1^ causes the terms to oscillate in sign, so
lim k→∞ (−1)k+^
2 k + 1 4 k + 3
is undefined.
The series diverges by the Zero Limit Test.
- Show that the Ratio Test always fails for a p-series
∑^ ∞
k=
kp^
, where p > 0.
The ratio of successive terms is
lim k→∞
ak+ ak
= lim k→∞
(k + 1)p 1 kp
= lim k→∞
k k + 1
)p = 1.
Since the limit is 1, the Ratio Test fails.
Remark. This problem also shows that it’s useless to apply the Ratio Test to series where the k-th term is a rational function of k. For example, it’s useless to apply the Ratio Test to
∑^ ∞
k=
k^2 + 5 k^4 + 3
since for large k,
k^2 + 5 k^4 + 3
k^2
, and the series is essentially a p-series.
Happiness depends upon ourselves. - Aristotle
©^ c2008 by Bruce Ikenaga 14
