Math 101—April 2015 Final Examination Solutions
Problems 1–2 are answer-only questions: the correct answers in the given boxes earns full credit,
and no partial credit is given.
1a. [2 pts] Only one of the following statements is always true; determine which one is true.
(Assume f (x) and g(x) are continuous functions.)
Answer:
E
Z −2
Z 2
A:
f (x)dx = −
f (x)dx.
−3
3
Z 1
Z 1
Z 1
B:
f (x) · g(x) dx =
f (x) dx ·
g(x) dx.
0Z
0
0
Z ∞
∞
C: If
f (x) dx converges and g(x) ≥ f (x) ≥ 0 for all x, then
g(x) dx converges.
1
D: If
∞
X
(−1)n+1 bn converges, then
n=1
∞
X
1
bn also converges.
n=1
E: When f (x) is positive and concave up, any Trapezoid Rule approximation for
Z b
will be an upper estimate for
f (x) dx.
Z
b
f (x) dx
a
a
Solution.
A: is false. For example if f (x) = 0 for x < 0 and f (x) = 1 for x ≥ 0, then
R −2
R2
f (x)dx = 0 and 3 f (x)dx = 1.
−3
R1
R1
B: is false. For example, if f (x) = g(x) = x, then 0 f (x) · g(x) dx = 0 x2 dx = 31 and
R1
R1
R1
R1
f (x) dx · 0 g(x) dx = 0 x dx · 0 x dx = 21 · 12
0
R∞
−x
C: is
false.
For
example
if
f
(x)
=
e
and
g(x)
=
1
then
f (x) dx converges but
1
R∞
g(x)
dx
diverges.
1
P
P
n+1 1
n+1
D: is false. For example if bn = n1 , then ∞
bn = ∞
converges,
n=1 (−1)
n=1 (−1)
n
P∞ 1
but n=1 n diverges.
E: is true. Because f (x) is positive and concave up, the graph of f (x) is always below
the top of the trapezoids used in the trapezoidal rule.
1b. [2 pts] For each integral, choose the substitution type that is most beneficial for evaluating
the integral. (Write F, G, or H in each box; each answer will be used exactly once.)
F: x = a sec θ
G: x = a sin θ
H: x = a tan θ
Z
2
2x
√
(i)
dx
Answer:
F
9x2 − 16
Z
x4 − 3
√
dx
Answer:
(ii)
G
1 − 4x2
Z
(iii)
(25 + x2 )−5/2 dx
Answer:
H
Math 101—April 2015 Final Examination
page 2
Solution.
(i) The substitution x = 34 sec θ, combined with sec2 θ − 1 = tan2 θ eliminates the square
√
root 9x2 − 16.
(ii) The substitution x = 12 sin θ, combined with 1 − sin2 θ = cos2 θ eliminates the square
√
root 1 − 4x2 .
(iii) The substitution
x = 5 tan θ, combined with 1 + tan2 θ = sec2 θ eliminates the square
√
root 25 + x2 .
Z 1
1c. [2 pts] For which values of the constant q does the integral
xq dx converge?
0
Answer:
L
J: It converges exactly when q < −1.
K: It converges exactly when q ≤ −1.
L: It converges exactly when q > −1.
M: It converges exactly when q ≥ −1.
P: It converges for all q.
Solution. The improper integral
 q+1
1−a

Z 1
Z 1
 q+1
xq dx = lim
xq dx = lim − log a
a→0+
a→0+ 
0
a
 1−aq+1
q+1
1d. [2 pts] Calculate
Z
7
−7
 
1
if q > −1 
 
 q+1 if q + 1 > 0
if q = −1 = +∞ if q = −1
 
if q < −1  +∞ if q + 1 < 0
x3 + sin(2x) + 1 dx. Simplify your answer completely.
Answer:
14
Solution. Both x3 and sin(2x) are odd functions so that
Z 7
Z 7
3
x dx =
sin(2x) dx = 0
−7
−7
2a. [2 pts] Which of the following integrals equals
A:
B:
C:
Z
Z
Z
2
2
3u f (u) du
1
D:
1
√
3
2
3u2 f (u) du
E:
1
Z
2
2
3u f (u) du
F:
Z
1
8
2
f (x3 ) dx?
Answer:
1
f (u)
du
3u2/3
√
3
1
8
1
Z
Z
G:
1
2
f (u)
du
3u2/3
f (u)
du
3u2/3
Z
H:
Z
2
f (u) du
√
3
2
f (u) du
1
J:
Z
8
f (u) du
1
Solution. Making the substitution u = x3 , du = 3x2 dx = 3u2/3 dx gives
Z 2
Z 23
du
3
f (x ) dx =
f (u) 2/3
3u
1
13
F
Math 101—April 2015 Final Examination
page 3
4
K : x3 − 2x7 + 4x11 − 8x15 + · · ·
x
? Answer:
1 − 2x3
Q : x4 − 2x7 + 4x10 − 8x13 + · · ·
M : x3 + 2x7 + 4x11 + 8x15 + · · ·
T : x4 + 2x7 + 4x10 + 8x13 + · · ·
2b. [2 pts] Which of the following Maclaurin series equals
L : x3 − 21 x7 + 41 x11 − 81 x15 + · · ·
P : x3 + 12 x7 + 41 x11 + 18 x15 + · · ·
Solution. Applying
1
1−r
T
S : x4 − 12 x7 + 41 x10 − 81 x13 + · · ·
U : x4 + 21 x7 + 14 x10 + 18 x13 + · · ·
= 1 + r + r 2 + r 3 + · · · with r = 2x3 gives
x4
4
3
3 2
3 3
=
x
1
+
(2x
)
+
(2x
)
+
(2x
)
+
·
·
·
1 − 2x3
= x4 + 2x7 + 4x10 + 8x13 + · · ·
2c. [4 pts] For each of the following series, choose the appropriate statement. (Write the
appropriate two-letter answer in each box; each answer will be used at most once.)
R∞
CI: Converges by Integral Test with 1 lnxx dx
CT: Converges by Test for Divergence P
1
CC: Converges by Comparison Test with ∞
n=1 n3
CA: Converges by Alternating Series Test
P
1
CL: Converges by Limit Comparison
with ∞
n=1 n2
R ∞Test
ln x
DI: Diverges by Integral Test with 1 x dx
DT: Diverges by Test for Divergence P
√1
DC: Diverges by Comparison Test with ∞
n=1 n
DA: Diverges by Alternating Series Test
P
π
DL: Diverges by Limit Comparison Test with ∞
n=1 n2
(i)
(ii)
∞
X
ln n
n=1
∞
X
n=1
(iii)
∞
X
Answer:
DI
4n − 1
(2n2 + 1)2
Answer:
CC
(−1)n n
Answer:
DT
n
n=1
(iv)
∞
X
arctan n
n=1
n2
Answer:
CL
Math 101—April 2015 Final Examination
page 4
Problems 3–6 are short-answer questions: put a box around your final answer, but you must include
the correct accompanying work to receive full credit.
√
3a. [3 pts] Calculate the area of the region enclosed by y = 2x and y = x + 1.
Solution. The two curves cross when x = 0 (y = 1 for both curves) and x = 1 (y = 2 for
x
both curves). Since 2x = (elog 2 ) = ex log 2 , the area is
Z 1
h
√
1 x i1
2 3/2
x log 2
2
dx = x + x −
Area =
(1 + x) − e
0
3
log 2
0
= 1+
3b. [3 pts] Evaluate
Z
2
5
1
1
−
[2 − 1] = −
3 log 2
3 log 2
tan3 x sec5 x dx.
Solution. Substituting u = cos x, du = − sin x dx, sin2 x = 1 − cos2 x = 1 − u2 , gives
Z
Z
Z
Z
(1 − cos2 x) sin x
1 − u2
sin3 x
3
5
dx
=
dx
=
−
du
tan x sec x dx =
cos8 x
cos8 x
u8
h u−7 u−5 i
1
1
=−
+ C = sec7 x − sec5 x + C
−
−7
−5
7
5
Solution. Alternatively, substituting u = sec x, du = sec x tan x dx, tan2 x = sec2 x−1 =
u2 − 1, gives
Z
Z
Z
3
5
2
4
tan x sec x dx = tan x sec x (tan x sec x) dx = (u2 − 1)u4 du
=
1
u5 i
1
+ C = sec7 x − sec5 x + C
−
7
5
7
5
h u7
4a. [3 pts] Find the average value of the function y = x2 ln x on the interval 1 ≤ x ≤ e.
Solution.
3
x=e
3
Z e
1
1
1
x
x3
e
e3 1
2
Ave =
=
x ln x dx =
ln x −
− +
e−1 1
e−1 3
9 x=1 e − 1 3
9
9
1
2 3 1
e +
=
e−1 9
9
R 2
3
3
The indefinite integral x ln x dx = x3 ln x − x9 + C was guessed and then verified by
3
d
x3
checking that dx
ln x − x9 = x2 ln x. The same indefinite integral can be found by
3
3
using integration by parts with u = ln x, dv = x2 dx, v = x3 .
Math 101—April 2015 Final Examination
page 5
Z
1
dx.
4b. [3 pts] Calculate
4
x + x2
Solution. We’ll first do a partial fractions expansion. The sneaky way is to temporarily
rename x2 to y. Then x4 + x2 = y 2 + y and
1
1
1
1
=
= −
4
2
x +x
y(y + 1)
y y+1
Now we restore y to x2 giving
Z
Z h
1
1
1 i
1
dx = − − arctan x + C
dx
=
−
4
2
2
2
x +x
x
x +1
x
∞
X
(−1)n
is absolutely convergent, conditionally
5a. [3 pts] Determine whether the series
9n
+
5
n=1
convergent, or divergent; justify your answer.
P
(−1)n
Solution. The series ∞
converges by the alternating series test. On the other
n=1
P∞
P∞ (−1)n 9n+5
1
diverges by the limiting comparison test with
hand the series n=1 9n+5 = n=1 9n+5
1
bn = n . So the answer is conditionally convergent .
∞
X
(−1)n+1
converges to some number S (you don’t have to prove
5b. [3 pts] The series
(2n + 1)2
n=1
this). According to the Alternating Series Estimation Theorem, what is the smallest value
1
away from S? For this value
of n for which the nth partial sum of the series is at most 100
of n, write out the nth partial sum of the series.
Solution. The error introduced when we approximate S by Sn lies between 0 and the first
n+2
term dropped, which is (−1)
. So we need the smallest positive integer n obeying
(2n+3)2
1
7
1
≤
⇐⇒ (2n + 3)2 ≥ 100 ⇐⇒ 2n + 3 ≥ 10 ⇐⇒ n ≥
2
(2n + 3)
100
2
1
1
1
1
− 2+ 2− 2
2
3
5
7
9
Z
6a. [3 pts] Find the Maclaurin series for x4 arctan(2x) dx.
So we need n = 4 and then S4 =
Solution. Recall that
arctan(y) =
∞
X
n=0
(−1)n
y 2n+1
2n + 1
Setting y = 2x, we have
Z 2n+1 2n+5
Z
Z
∞
∞
2n+1
X
X
2
x
n
4 (2x)
4
n
dx =
(−1)
dx
x
x arctan(2x) dx =
(−1)
2n + 1
2n + 1
n=0
n=0
∞
X
∞
X
22n+1 x2n+6
22n x2n+6
(−1)
=
+C =
(−1)n
+C
(2n
+
1)(2n
+
6)
(2n
+
1)(n
+
3)
n=0
n=0
n
Math 101—April 2015 Final Examination
page 6
P∞
6b. [3 pts] Suppose that you have a sequence {bn } such that the series n=0 (1−bn ) converges.
Using the tests we’ve learned in class, prove that the radius of convergence of the power
∞
X
series
bn xn is equal to 1.
n=0
P
Solution. By the divergence test, the fact that ∞
guarantees that
n=0 (1 − bn ) converges
P∞
limn→∞ bn = 1. So, P
by the limiting comparison test, the series n=0 bn xn converges if
n
and only if the series ∞
n=0 x converges, which is the case if and only if |x| < 1. So the
radius of convergence is indeed 1.
Problems 7–11 are long-answer: give complete arguments and explanations for all your calculations—
answers without justifications will not be marked.
Z x3 +1
3
3
et dt.
7. Define f (x) = x
0
(a) [3 pts] Find a formula for the derivative f ′ (x).
Solution. Using the product rule, followed by the chain rule, followed by the fundamental
theorem of calculus,
Z x3 +1
Z x3 +1
3
′
2
t3
3d
et dt
f (x) = 3x
e dt + x
dx 0
0
" Z
#
Z x3 +1
y
d
3
3
et dt + x3 3x2
= 3x2
et dt
dy
0
0
y=x3 +1
Z x3 +1
h
i
3
3
et dt + x3 3x2 ey
= 3x2
y=x3 +1
0
= 3x2
Z
x3 +1
0
2
= 3x
Z
3 3
3
et dt + x3 3x2 e(x +1)
x3 +1
3
et dt + 3x5 e(x
3 +1)3
0
(b) [3 pts] Find the equation of the tangent line to the graph of y = f (x) at x = −1. Simplify
your answer completely.
Solution. In general, the equation of the tangent line to the graph of y = f (x) at x = a is
y = f (a) + f ′ (a) (x − a)
Substituting in the given f (x) and
a3 + 1 = (−1)3 + 1 = 0
a = −1
f (−1) = (−1)
3
Z
0
0
3
et dt = 0
3
f ′ (−1) = 3(−1)2 × 0 + 3(−1)5 e0 = −3
the equation of the tangent line is y = −3(x + 1)
Math 101—April 2015 Final Examination
√
page 7
8. Let D be the region below the graph of the curve y = 9 − 4x2 and above the x-axis.
(a) [4 pts] Using an appropriate integral, find the area of the region D; simplify your answer
completely.
Solution. Notice that when x = 0, y = 3 and as x2 increases, y decreases until y hits zero
at x2 = 49 , i.e. at x = ± 23 . For x2 > 94 , y is not even defined.
√
As an aside, we can rewrite y = 9 − 4x2 as 4x2 + y 2 = 9, y ≥ 0, which is the top half
of the ellipse which passes through (±a, 0) and (0, ±b) with a = 23 and b = 3. The area of
the full ellipse is πab = 92 π. The area of D is half of that, which is 94 π. But we are told to
use an integral, so we will do so.
The area is
Z 3/2 √
9 − 4x2 dx
Area =
−3/2
We can evaluate this integral by substituting x = 23 sin θ, dx = 23 cos θ dθ and using
3
x = ± ⇐⇒ sin θ = ±1
2
π
π
So − 2 ≤ θ ≤ 2 and
Z π/2 p
Z π/2 q
2 3
3
3
cos θ dθ =
9 − 4 2 sin θ
9 − 9 sin2 θ cos θ dθ
Area =
2
2
−π/2
−π/2
Z π/2
Z π/2
9
cos(2θ) + 1
9
cos2 θ dθ
=
dθ
=
2 −π/2
2 −π/2
2
iπ/2
9 h sin(2θ)
9
=
= π
+θ
4
2
4
−π/2
(b) [3 pts] Find the centre of mass of the region D; simplify your answer completely. (Assume
it has constant density ρ.)
Solution. The region D is symmetric about the y axis. So the centre of mass
√lies on the y
axis. That is, x̄ = 0 . Since D has area A = 94 π, top equation y = T (x) = 9 − 4x2 and
bottom equation y = B(x) = 0, with x running from a = − 32 to b = 32 ,
Z b
Z 3/2
Z 3/2
1
2
4
2
2
2
ȳ =
T (x) − B(x) dx =
9 − 4x dx =
9 − 4x2 dx
2A a
9π −3/2
9π 0
4h
4 h 3 4 33 i
4 i3/2
4h 3
1i
=
=
9x − x3
9 −
9
=
−
9
0
9π
3
9π 2 3 23
9π 2
2
4
=
π
Math 101—April 2015 Final Examination
page 8
9.
(a) [3 pts] Find the solution of the differential equation
1+
p
y 2 − 4 ′ sec x
y =
that satisfies
tan x
y
y(0) = 2. You don’t have to solve for y in terms of x.
Solution. This is a separable differential equation.
p
p
1 + y 2 − 4 ′ sec x
y =
⇐⇒ y 1 + y 2 − 4 dy = sec x tan x dx
tan x
y
Z
Z
p
2
⇐⇒
y 1 + y − 4 dy = sec x tan x dx
⇐⇒
3/2
y2 1 2
= sec x + C
+ y −4
2
3
To determine C we set x = 0 and y = 2.
3/2
22 1 2
= sec 0 + C = 1 + C ⇒ C = 1
+ 2 −4
2
3
So the solution is
3/2
y2 1 2
= sec x + 1
+ y −4
2
3
(b) [4 pts] A sculpture, shaped like a pyramid 3 m high sitting on the ground, has been made
by stacking smaller and smaller iron plates on top of one another. The iron plate at height
z m above ground level is a square whose side length is (3 − z) m. All of the iron plates
started on the floor of a basement 2 m below ground level.
Write down an integral that represents the work, in newtons, it took to move all of the
iron from its starting position to its present position. Do not evaluate the integral. (You can
use 9.8 m/s2 for the force of gravity and 8000 kg/m3 for the density of iron.)
Solution. This question is sloppily worded in that the plates are to be “infinitesmally
thin”. Also the units of work are joules, not newtons. (Newtons are units of force.) The
plate at height z has
◦ has side length 3 − z and hence
◦ has area (3 − z)2 and hence
◦ has volume (3 − z)2 dz and hence
◦ has mass 8000(3 − z)2 dz and hence
◦ is subject to a gravitational force of 9.8 × 8000(3 − z)2 dz and hence
◦ requires work 9.8 × 8000(2 + z)(3 − z)2 dz to raise it from 2m below ground level to
zm above ground level.
So the total work is
Z 3
9.8 × 8000(2 + z)(3 − z)2 dz joules
0
Math 101—April 2015 Final Examination
page 9
2
2
10. Let R be the region inside the circle x + (y − 2) = 1. Let S be the solid obtained by rotating
R about the x-axis.
(a) [4 pts] Write down an integral representing the volume of S.
√
Solution. The top and
the
bottom
of
the
circle
have
equations
y
=
T
(x)
=
2
+
1 − x2
√
2
and y = B(x) = 2 − 1 − x , respectively.
y = T (x)
y = B(x)
y=0
x = −1
x=1
When R is rotated about the x–axis, the vertical strip of R in the figure above sweeps out
a washer with thickness dx, outer radius T (x) and inner radius B(x). This washer has
volume
√
π T (x)2 − B(x)2 dx = π T (x) + B(x) T (x) − B(x) dx = π × 4 × 2 1 − x2 dx
Hence the volume of the solid is
8π
Z
1
−1
√
1 − x2 dx
(b) [3 pts] Evaluate the integral you wrote down in part (a). (You may use any technique you
know, including a geometric
one, to do so.)
√
Solution. Since y = 1 − x2 is equivalent to x2 + y 2 = 1, y ≥ 0, the integral is 8π times
the area of the upper half of the circle x2 + y 2 = 1 and hence is 8π × 12 π12 = 4π 2 .
Math 101—April 2015 Final Examination
page 10
11.
(a) [4 pts] Find the value of the convergent series
∞ n+1
X
1
1
2
+
−
3n
2n − 1 2n + 1
n=2
(simplify your answer completely).
Solution. We split the sun into two parts. The first part is a geometric series with first
2+1
term a = 2 32 = 98 and ratio r = 32 .
∞
X
2n+1
n=2
The rest telescopes.
3n
=
a
8/9
8
=
=
1−r
1 − 2/3
3
n=2
∞ X
n=2
So
1
1
−
2n − 1 2n + 1
∞ n+1
X
2
n=2
3n
n=3
n=4
z }| { z }| { z }| {
1 1
1 1
1 1
1
=
+
+
+··· =
−
−
−
3 5
5 7
7 9
3
1
1
+
−
2n − 1 2n + 1
=
8 1
+ = 3
3 3
(b) [4 pts] Find the interval of convergence for the power series
∞
X
(x − 2)n
.
4/5 (5n − 4)
n
n=1
n
(x−2)
. Since
Solution. We first apply the ratio test with an = n4/5
(5n −4)
a (x − 2)n+1
n4/5 (5n − 4) n+1 lim = lim n→∞
n→∞ (n + 1)4/5 (5n+1 − 4)
an
(x − 2)n n4/5 (5n − 4)
|x − 2|
n→∞ (n + 1)4/5 (5n+1 − 4)
(1 − 4/5n )
= lim
|x − 2|
n→∞ (1 + 1/n)4/5 (5 − 4/5n )
|x − 2|
=
5
the series converges
< 5 and
diverges if |x − 2| > 5. When x − 2 = +5, the
P∞
P if |x −52|
n
1
=
series reduces to ∞
n=1 n4/5 (1−4/5n ) which diverges by the limit comn=1 n4/5 (5n −4)
P
(−5)n
1
parison test with bn = n4/5
. When x − 2 = −5, the series reduces to ∞
n=1 n4/5 (5n −4) =
P∞
(−1)n
n=1 n4/5 (1−4/5n ) which converges by the alternating series test. So the interval of conver= lim
gence is −3 ≤ x < 7 .