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Math 211- 4–30–
Review Problems for the Final
These problems are provided to help you study. The presence of a problem on this handout does not
imply that there will be a similar problem on the test. And the absence of a topic does not imply that it
won’t appear on the test.
- Find the area of the intersection of the interiors of the circles
x 2
2
- Does the series 1
2
converge absolutely, converge conditionally, or diverge?
- Find the sum of the series 5
9
- In each case, determine whether the series converges or diverges.
(a)
∑^ ∞
n=
n(ln n)^4 /^3
(b)
2 · 5 · · · · · (3n − 1)
1 · 5 · · · · · (4n − 3)
(c)
∑^ ∞
n=
n
)−n^2
.
(d)
(e)
∑^ ∞
n=
3 n 2
(f)
∑^ ∞
n=
5 + cos(e n )
n
- Find the values of x for which the series
∑^ ∞
n=
(n!)^2
(2n)!
(x − 5) n
converges absolutely.
- Compute the following integrals.
(a)
e x cos 2x dx.
(b)
x 2 √ 4 − x^2
dx.
(c)
5 x^2 − 6 x − 5
(x − 1)^2 (x + 2)
dx.
(d)
(sin 4x) 3 (cos 4x) 2 dx.
(e)
(sin 4x) 2 (cos 4x) 2 dx.
(f)
(− 3 − 4 x − x^2 )^3 /^2
dx.
- Let R be the region bounded above by y = x + 2, bounded below by y = −x^2 , and bounded on the sides
by x = −2 and by the y-axis. Find the volume of the solid generated by revolving R about the line x = 1.
- Compute lim x→∞
x^2 + 8x − x
- Compute lim x→∞
x
) 3 x
.
- If x = t + e t and y = t + t 3 , find
dy
dx
and
d 2 y
dx^2
at t = 1.
- (a) Find the Taylor expansion at c = 1 for e^2 x.
(b) Find the Taylor expansion at c = 1 for
3 + x
. What is the interval of convergence?
- Find the area of the region which lies between the graphs of y = x 2 and y = x + 2, from x = 1 to x = 3.
- Find the area of the region between y = x + 3 and y = 7 − x from x = 0 to x = 3.
- The base of a solid is the region in the x-y-plane bounded above by the curve y = e x , below by the x-axis,
and on the sides by the lines x = 0 and x = 1. The cross-sections in planes perpendicular to the x-axis are
squares with one side in the x-y-plane. Find the volume of the solid.
- Find the interval of convergence of the power series
∑^ ∞
n=
(x − 3)n
n(2n)^3
- Find the slope of the tangent line to the polar curve r = sin 2θ at θ =
π
6
- A tank built in the shape of the bottom half of a sphere of radius 2 feet is filled with water. Find the
work done in pumping all the water out of the top of the tank.
- Let
x =
t 2 , y = t −
t 3 .
Find the length of the arc of the curve from t = −2 to t = 2.
- Find the area of the surface generated by revolving y =
x^3 , 0 ≤ x ≤ 2, about the x-axis.
- (a) Convert (x − 3) 2
- (y + 4) 2 = 25 to polar and simplify.
(b) Convert r = 4 cos θ − 6 sin θ to rectangular and describe the graph.
- Find the area of the region inside the cardioid r = 1 + cos θ and outside the circle r = 3 cos θ.
The absolute value series is ∑∞
n=
n 2
n^3 + 1
Since n^2
n^3 + 1
n
for large values of n, I’ll compare the series to
n=
n
lim n→∞
n^2
n^3 + 1 1
n
= lim n→∞
n 3
n^3 + 1
The limit is finite ( 6 = ∞) and positive (> 0). The harmonic series
n=
n
diverges. By Limit Compar-
ison, the series
n=
n 2
n^3 + 1
diverges. Hence, the original series does not converge absolutely.
Returning to the original series, note that it alternates, and
lim n→∞
n 2
n^3 + 1
Let f (n) =
n^2
n^3 + 1
. Then
f
′ (n) =
n(2 − n 3 )
(1 + n^3 )^2
for n > 1. Therefore, the terms of the series decrease for n ≥ 2, and I can apply the Alternating Series Rule
to conclude that the series converges. Since it doesn’t converge absolutely, but it does converge, it converges
conditionally.
- Find the sum of the series 5
9
- In each case, determine whether the series converges or diverges.
(a)
∑^ ∞
n=
n(ln n)^4 /^3
Apply the Integral Test. The function f (n) =
n(ln n)^4 /^3
is positive and continuous on the interval
[2, +∞).
Since
f
′ (n) = −
3 n^2 (ln n)^7 /^3
n^2 (ln n)^4 /^3
it follows that f ′ (n) < 0 for n ≥ 2. Hence, f decreases on the interval [2, +∞). The hypotheses of the
Integral Test are satisfied.
Compute the integral:
∫ (^) ∞
2
n(ln n)^4 /^3
dn = lim p→∞
∫ (^) p
2
n(ln n)^4 /^3
dn =
lim p→∞
[
(ln n)^1 /^3
]p
2
= −3 lim p→∞
(ln p)^1 /^3
(ln 2)^1 /^3
(ln 2)^1 /^3
(To do the integral, I substituted u = ln n, so du =
n
dn.)
Since the integral converges, the series converges, by the Integral Test.
(b)
2 · 5 · · · · · (3n − 1)
1 · 5 · · · · · (4n − 3)
Apply the Ratio Test. The nth^ term of the series is
an =
2 · 5 · · · · · (3n − 1)
1 · 5 · · · · · (4n − 3)
so the (n + 1)-st term is
an+1 =
2 · 5 · · · · · (3n − 1) · (3(n + 1) − 1)
1 · 5 · · · · · (4n − 3) · (4(n + 1) − 3)
Hence,
an+
an
2 · 5 · · · · · (3n − 1) · (3(n + 1) − 1)
1 · 5 · · · · · (4n − 3) · (4(n + 1) − 3)
1 · 5 · · · · · (4n − 3)
2 · 5 · · · · · (3n − 1)
3(n + 1) − 1)
4(n + 1) − 3)
3 n + 2
4 n + 1
The limiting ratio is
lim n→∞
3 n + 2
4 n + 1
The limit is less than 1, so the series converges, by the Ratio Test.
(c)
∑^ ∞
n=
n
)−n^2
.
Apply the Root Test.
a
1 /n n =
n
)−n
.
The limit is
lim n→∞
n
)−n
= lim n→∞
n
)n}− 1
=
lim n→∞
n
)n}− 1
= e − 1 .
Since e − 1 =
e
< 1, the series converges, by the Root Test.
(d)
Since
lim n→∞
2 + 3n
3 + 5n
it follows that limn→∞ an is undefined — the values oscillate, approaching ±
. Since, in particular, the
limit is nonzero, the series diverges, by the Zero Limit Test.
(a)
e
x cos 2x dx.
d
dx
dx
− e x 1 2
sin 2x
ց
cos 2x
∫
e x cos 2x dx =
e x sin 2x +
e x cos 2x −
e x cos 2x dx,
e x cos 2x dx =
e x sin 2x +
e x cos 2x,
e x cos 2x dx =
e x sin 2x +
e x cos 2x + C.
(b)
x^2 √ 4 − x^2
dx.
x^2 √ 4 − x^2
dx =
4(sin θ)^2 √ 4 − 4(sin θ)^2
2 cos θ dθ =
4(sin θ)^2 √ 4(cos θ)^2
2 cos θ dθ = 4
(sin θ) 2 dθ =
[x = 2 sin θ, dx = 2 cos θ dθ]
(1 − cos 2θ) dθ = 2
θ −
sin 2θ
- C = 2 (θ − sin θ cos θ) + C = 2 sin
− 1 x 2
x
4 − x^2 + C.
x
q
2
4 - x^2
(c)
5 x^2 − 6 x − 5
(x − 1)^2 (x + 2)
dx.
5 x 2 − 6 x − 5
(x − 1)^2 (x + 2)
a
x − 1
b
(x − 1)^2
c
x + 2
5 x 2 − 6 x − 5 = a(x − 1)(x + 2) + b(x + 2) + c(x − 1) 2 .
Setting x = 1 gives −6 = 3b, so b = −2.
Setting x = −2 gives 27 = 9c, so c = 3. Therefore,
5 x 2 − 6 x − 5 = a(x − 1)(x + 2) − 2(x + 2) + 3(x − 1) 2 .
Setting x = 0 gives −5 = − 2 a − 4 + 3, so a = 2.
Thus,
5 x 2 − 6 x − 5
(x − 1)^2 (x + 2)
dx =
∫ (^
x − 1
(x − 1)^2
x + 2
dx = 2 ln |x − 1 | +
x − 1
(d)
(sin 4x)
3 (cos 4x)
2 dx.
∫
(sin 4x) 3 (cos 4x) 2 dx =
(sin 4x) 2 (cos 4x) 2 (sin 4x dx) =
1 − (cos 4x) 2
(cos 4x) 2 (sin 4x dx) =
[
u = cos 4x, du = −4 sin 4x dx, dx =
du
−4 sin 4x
]
(1 − u 2 )u 2 (sin 4x)
du
−4 sin 4x
(u 4 − u 2 ) du =
u 5 −
u 3
+ C =
(cos 4x) 5 −
(cos 4x) 3
+ C.
(e)
(sin 4x) 2 (cos 4x) 2 dx.
∫
(sin 4x) 2 (cos 4x) 2 dx =
(1 − cos 8x) ·
(1 + cos 8x) dx =
1 − (cos 8x) 2
dx =
(sin 8x) 2 dx =
(1 − cos 16x) dx =
x −
sin 16x
+ C.
(f)
(− 3 − 4 x − x^2 )^3 /^2
dx.
I need to complete the square. Note that
= −2 and (−2)^2 = 4. Then
− 3 − 4 x − x 2 = −(x 2
- 4x + 3) = −(x 2
- 4x + 4 − 1) = −
[
(x + 2) 2 − 1
]
= 1 − (x + 2) 2 .
So ∫ 1
(− 3 − 4 x − x^2 )^3 /^2
dx =
(1 − (x + 2)^2 )^3 /^2
dx =
(1 − (sin θ)^2 )^3 /^2
(cos θ dθ) =
(cos θ)^3
(cos θ dθ) =
[x + 2 = sin θ, dx = cos θ dθ]
x + 2
1
1 - (x + 2)^2
q
(cos θ)^2
dθ =
(sec θ) 2 dθ = tan θ + C =
x + 2 √ − 3 − 4 x − x^2
+ C.
- Let R be the region bounded above by y = x + 2, bounded below by y = −x^2 , and bounded on the sides
by x = −2 and by the y-axis. Find the volume of the solid generated by revolving R about the line x = 1.
dx
r = 1 - x
-x 1
x = 1
h
h = (x + 2) + x^2
When t = 1,
dy
dx
1 + e
d 2 y
dx^2
d
dx
dy
dx
dt
dx
d
dt
dy
dx
d
dt
dy
dx
dx
dt
d
dt
1 + 3t^2
1 + et 1 + et^
(1 + e t )(6t) − (1 + 3t 2 )(e t )
(1 + et)^2
1 + et^
(1 + e t )(6t) − (1 + 3t 2 )(e t )
(1 + et)^3
When t = 1,
d^2 y
dx^2
6 + 2e
(1 + e)^3
- (a) Find the Taylor expansion at c = 1 for e^2 x.
e 2 x = e 2(x−1)+ = e 2 e 2(x−1) = e 2
1 + 2(x − 1) +
2 (x − 1) 2
3 (x − 1) 3
(b) Find the Taylor expansion at c = 1 for
3 + x
. What is the interval of convergence?
3 + x
4 + (x − 1)
x − 1
4
x − 1
4
x − 1
4
x − 1
4
x − 1
4
The series converges for − 1 <
x − 1
4
< 1, i.e. for − 3 < x < 5.
- Find the area of the region which lies between the graphs of y = x^2 and y = x + 2, from x = 1 to x = 3.
1 2 3 4
2
4
6
8
10
2
4
6
8
As the picture shows, the curves intersect. Find the intersection point:
x 2 = x + 2, x 2 − x − 2 = 0, (x − 2)(x + 1) = 0, x = 2 or x = − 1.
On the interval 1 ≤ x ≤ 3, the curves cross at x = 2. I’ll use vertical rectangles. From x = 1 to x = 2,
the top curve is y = x + 2 and the bottom curve is y = x 2
. From x = 2 to x = 3, the top curve is y = x 2
and the bottom curve is y = x + 2. The area is
A =
1
(x + 2) − x 2
dx +
2
x 2 − (x + 2)
dx = 3.
- Find the area of the region between y = x + 3 and y = 7 − x from x = 0 to x = 3.
-1 1 2 3 4
3
4
5
6
7
3
4
5
6
7
As the picture shows, the curves intersect. Find the intersection point:
x + 3 = 7 − x, 2 x = 4, x = 2.
I’ll use vertical rectangles. From x = 0 to x = 2, the top curve is y = 7 − x and the bottom curve is
y = x + 3. From x = 2 to x = 3, the top curve is y = x + 3 and the bottom curve is y = 7 − x. The area is
∫ (^2)
0
((7 − x) − (x + 3)) dx +
2
((x + 3) − (7 − x)) dx =
0
(4 − 2 x) dx +
2
(2x − 4) dx =
[
4 x − x 2
] 2
0
[
x 2 − 4 x
] 3
2
- The base of a solid is the region in the x-y-plane bounded above by the curve y = e x , below by the x-axis,
and on the sides by the lines x = 0 and x = 1. The cross-sections in planes perpendicular to the x-axis are
squares with one side in the x-y-plane. Find the volume of the solid.
y
(0,0)
x
The volume is
V =
0
(e x ) 2 dx =
0
e 2 x dx =
[
e 2 x
] 1
0
(e 2 − 1) ≈ 3. 19453.
I’ve drawn the tank in cross-section as a semicircle of radius 2 extending from y = −2 to y = 0. Divide the volume of water up into circular slices. The radius of a slice is r =
4 − y^2 , so the volume of
a slice is dV = πr 2 dy = π(4 − y 2 ) dy. The weight of a slice is 62. 4 π(4 − y 2 ) dy, where I’m using 62.4 pounds
per cubic foot as the density of water.
To pump a slice out of the top of the tank, it must be raised a distance of −y feet. (The “-” is necessary
to make y positive, since y is going from −2 to 0.)
The work done is
W =
− 2
- 4 π(−y)(4 − y 2 ) dy = 62. 4 π
− 2
(y 3 − 4 y) dy = 62. 4 π
[
y 4 − 2 y 2
] 0
− 2
6 π ≈ 784 .14153 foot − pounds.
Let
x =
t
2 , y = t −
t
3 .
Find the length of the arc of the curve from t = −2 to t = 2.
dx
dt
3 t and
dy
dt
t 2 ,
so
( dx
dt
dy
dt
= 3t 2
t 2
= 3t 2
t 2
t 4 = 1 +
t 2
t 4 =
t 2
Therefore, (^) √ ( dx
dt
dy
dt
t 2 .
The length is ∫ (^2)
− 2
t
2
dt =
[
t +
t
3
] 2
− 2
- Find the area of the surface generated by revolving y =
x 3 , 0 ≤ x ≤ 2, about the x-axis.
0
1
2
0
0
1
0
1
0
The derivative is
dy
dx
= x 2 , so
dy
dx
x^4 + 1.
The curve is being revolved about the x-axis, so the radius of revolution is R = y =
x^3. The area of
the surface is
S =
0
2 π
x 3
x^4 + 1 dx =
2 π
3
1
u 1 / 2 · x 3
du
4 x^3
π
6
1
u 1 / 2 du =
π
6
[
u 3 / 2
] 17
1
[
u = x 4
du
4 x^3
; x = 0, u = 1, x = 2, u = 17
]
π
9
3 / 2 − 1
- (a) Convert (x − 3) 2
- (y + 4) 2 = 25 to polar and simplify.
(x − 3) 2
- (y + 4) 2 = 25, x 2 − 6 x + 9 + y 2
- 8y + 16 = 25, x 2
- y 2 = 6x − 8 y,
r 2 = 6r cos θ − 8 r sin θ, r = 6 cos θ − 8 sin θ.
(b) Convert r = 4 cos θ − 6 sin θ to rectangular and describe the graph.
r = 4 cos θ − 6 sin θ, r 2 = 4r cos θ − 6 r sin θ, x 2
- y 2 = 4x − 6 y, x 2 − 4 x + y 2
- 6y = 0,
x 2 − 4 x + 4 + y 2
- 6y + 9 = 13, (x − 2) 2
- (y + 3) 2 = 13.
-1 1 2 3 4 5
The graph is a circle of radius
13 centered at (2, −3).
- Find the area of the region inside the cardioid r = 1 + cos θ and outside the circle r = 3 cos θ.
0.5 1 1.5 2 2.5 3
-1.
-0.
1
q=p/
r=1 + cosq
r=3 cosq
