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APPM 1360 Final Exam Fall 2013
On the front of your bluebook, please write: a grading key, your name, student ID, and section andinstructor. This exam is worth 150 points and has 9 questions.
• Show all work! Answers with no justification will receive no points.
• Please begin each problem on a new page.
• No notes, calculators, or electronic devices are permitted.
1. (20 points) Evaluate the following:
(a)∫
3
(x− 1)(x2 + 2)dx
(b) Solve for y in the differential equation:dy
dx= xy sinx with initial value y(π) = 1.
Solution:
(a) Use partial fractions.∫3
(x− 1)(x2 + 2)dx =
∫ (A
x− 1+Bx+ C
x2 + 2
)dx
A(x2 + 2) + (Bx+ C)(x− 1) = 3
(A+B)x2 + (−B + C)x+ (2A− C) = 3
A+B = 0, −B + C = 0, 2A− C = 3
A = 1, B = −1, C = −1
∫ (1
x− 1− x+ 1
x2 + 2
)dx =
∫ (1
x− 1− x
x2 + 2︸ ︷︷ ︸u=x2+2
− 1
x2 + 2
)dx
= ln |x− 1| − 1
2ln∣∣x2 + 2
∣∣− 1√2arctan
(x√2
)+ C
(b) This is a separable equation.
dy
dx= xy sinx ⇒
∫dy
y=
∫x︸︷︷︸u
sinx dx︸ ︷︷ ︸dv
Use integration by parts with u = x, du = dx, and dv = sinx dx, v = − cosx.
ln |y| = −x cosx+
∫cosx dx = −x cosx+ sinx+ C
|y| = e−x cosx+sinx+C
Plug in the initial value (π, 1).
1 = eπ+C ⇒ C = −π
y = e−x cosx+sinx−π
2. (14 points) Consider the region shown at right. Set up, butdo not evaluate, the integrals to calculate:
(a) The volume generated by rotating this region about theline x = 3.
(b) Consider the upper curve y = x2/2, defined on theinterval shown on the graph. Find the surface area ofthe solid generated by rotating the curve about they-axis. (The lower curve is not relevant for this part.)
y �
x2
2
y �
x4
4-
x2
2
x
y
Solution:
(a) The curves intersect at (0, 0) and (2, 2). By the Shell Method,
V =
∫2πrh dx =
∫ 2
02π(3− x)
(x2
2−(x4
4− x2
2
))dx
(b)
S =
∫2πr
√1 + (dy/dx)2 dx =
∫ 2
02πx
√1 + x2 dx
3. (24 points) Do the following converge or diverge? Explain your work and name any test or theoremyou use.
(a)∫ ∞0
dx
1 + ex(b){(lnm)2
m
}∞m=15
(c)∞∑n=3
√n+ 2
n− 2
Solution:
(a) Since 0 <1
1 + ex<
1
exand
∫ ∞0
dx
exis convergent, by the Comparison Theorem
∫ ∞0
dx
1 + ex
also is convergent .
Alternate solution:∫ ∞0
dx
1 + ex= lim
t→∞
∫ t
0
dx
1 + ex
Let u = 1 + ex, du = ex dx.
= limt→∞
∫ 1+et
2
du
u(u− 1)= lim
t→∞
∫ 1+et
2
(1
u− 1− 1
u
)du = lim
t→∞
[ln |u− 1| − ln |u|
]1+et2
= limt→∞
[ln
∣∣∣∣u− 1
u
∣∣∣∣]1+et2
= limt→∞
(ln
et
1 + et− ln
1
2
)= ln 1 + ln 2 = ln 2
The integral is convergent .
(b)
limm→∞
am = limm→∞
(lnm)2
m
L′H= lim
m→∞
2 lnm
m
L′H= lim
m→∞
2
m= 0
The sequence converges to 0.
(c) Since√n+ 2
n− 2>
√n
n=
1√n
and∞∑n=1
1√n
is a divergent p-series, by the Comparison Test
∞∑n=3
√n+ 2
n− 2also is divergent .
Alternate solution: Use the Limit Comparison Test with bn = 1/√n.
limn→∞
∣∣∣∣anbn∣∣∣∣ = lim
n→∞
∣∣∣∣√n+ 2
n− 2·√n
1
∣∣∣∣ = limn→∞
√n2 + 2n
n2 − 2n+ 4= 1
Since∞∑n=1
1√n
is a divergent p-series,∞∑n=3
√n+ 2
n− 2also is divergent .
4. (16 points) Find the values of x for which the following series converge absolutely, converge condi-
tionally, or diverge: (a)∞∑n=2
(x+ 3)n
n2 − 1(b)
∞∑n=0
3n
n!x2n
Solution:
(a) Use the Ratio Test.
limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = limn→∞
∣∣∣∣ (x+ 3)n+1
(n+ 1)2 − 1· n
2 − 1
(x+ 3)n
∣∣∣∣ = |x+ 3| < 1
The series is absolutely convergent on −4 < x < −2. Now check the endpoints. When
x = −2, the series∞∑n=2
1
n2 − 1converges by the Limit Comparison Test with the convergent
p-series∞∑n=1
1
n2. When x = −4, the series
∞∑n=2
(−1)n
n2 − 1converges by the Alternating Series Test
since1
(n+ 1)2 − 1<
1
n2 − 1and lim
n→∞
1
n2 − 1= 0.
Therefore the series∞∑n=2
(x+ 3)n
n2 − 1is absolutely convergent for −4 ≤ x ≤ −2 , is not condi-
tionally convergent for any x, and is divergent for x < −4, x > −2 .
(b) Use the Ratio Test.
limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = limn→∞
∣∣∣∣3n+1x2n+2
(n+ 1)!· n!
3nx2n
∣∣∣∣ = limn→∞
∣∣∣∣ 3x2
n+ 1
∣∣∣∣ = 0
Since the ratio is 0, the series is absolutely convergent for all x , and not conditionally conver-gent or divergent for any x.
5. (14 points) Let a1 = 1, a2 =1 + 1/4
1 + 1/2, a3 =
1 + 1/4 + 1/42
1 + 1/2 + 1/22, a4 =
1 + 1/4 + 1/42 + 1/43
1 + 1/2 + 1/22 + 1/23, . . ..
(a) Does the sequence {an} converge? If yes, find its limit. Explain.
(b) Does∞∑n=1
an converge? Explain your work and state any test or theorem you use.
Solution:
(a) an is the quotient of geometric series. The numerator has ratio 1/4 and the denominator hasratio 1/2. The sum of an infinite geometric series is S = a/(1− r).
limn→∞
an = limn→∞
1 + 14 + 1
42+ · · ·+ 1
4n−1
1 + 12 + 1
22+ · · ·+ 1
2n−1
=limn→∞
(1 + 1
4 + 142
+ · · ·+ 14n−1
)limn→∞
(1 + 1
2 + 122
+ · · ·+ 12n−1
) =1/(1− 1
4)
1/(1− 12)
=2
3
(b) By the Test for Divergence, since limn→∞
an 6= 0, the series diverges .
6. (20 points)
(a) Find a Maclaurin series for x sin(x2).
(b) Use your answer from part (a) to find a series for∫x sin(x2) dx.
(c) Use your answer from part (b) to find the 8th order Taylor polynomial T8(x) for the integral.
(d) Use T8(x) to approximate the value of∫ 1
0x sin(x2) dx.
(e) Is your approximation an overestimate or underestimate? Explain your reasoning.
Solution:
(a) sinx =∞∑n=0
(−1)n x2n+1
(2n+ 1)!⇒ x sin(x2) =
∞∑n=0
(−1)n x(x2)2n+1
(2n+ 1)!=
∞∑n=0
(−1)nx4n+3
(2n+ 1)!
(b)∫x sin(x2) dx = C +
∞∑n=0
(−1)nx4n+4
(2n+ 1)!(4n+ 4)
(c) T8(x) = C +x4
4− x8
48
(d)∫ 1
0x sin(x2) dx ≈ T8(1)− T8(0) =
1
4− 1
48=
11
48
(e) Because the series is alternating, the approximation is an underestimate .
7. (14 points)
(a) Sketch the graphs of r = −2 cos θ and r = 1.
(b) Find the area of the region inside r = −2 cos θ and outside r = 1.
Solution:
(a)
r@t_D := -2 Cos@tD;s@t_D := 1;Show@ParametricPlot@88r Cos@tD r@tD, r Sin@tD r@tD<, 8r Cos@tD s@tD, r Sin@tD s@tD<<,
8t, 0, 2 Pi<, 8r, 0, 1<,PlotRange Ø 88-2.5, 1.5<, 8-1.5, 1.5<<,AspectRatio Ø 3 ê 4,PlotStyle Ø [email protected], Blue<, 8Opacity@1D, White<<, Mesh Ø None,Ticks Ø 88-2, -1, 1<, 8-1, 1<<,Frame Ø None,AxesLabel Ø 8x, y<,AxesStyle Ø 88Arrowheads@AutomaticD, [email protected]<,
8Arrowheads@AutomaticD, [email protected]<<,LabelStyle Ø 24,PlotStyle Ø [email protected], Black<, [email protected], Black<<D,
PolarPlot@8-2 Cos@tD, 1<, 8t, 0, 2 Pi<,PlotStyle Ø [email protected], Black<, [email protected], Black<<,Mesh Ø NoneD
D
68 1360-exams-fall13.nb
(b) We wish to find the area of the shaded region. The curve r = −2 cos θ (for θ = 0 to π) andcurve r = 1 (for θ = 0 to 2π) intersect at −2 cos θ = 1 ⇒ θ = 2π/3. We double the area ofthe shaded region above the x-axis.
A = 2
∫1
2
(r21 − r22
)dθ = 2
∫ π
2π/3
1
2
((−2 cos θ)2 − 12
)dθ =
∫ π
2π/3
(4 cos2 θ − 1
)dθ
=
∫ π
2π/3(2(1 + cos 2θ)− 1) dθ =
∫ π
2π/3(1 + 2 cos 2θ) dθ
=
[θ + sin 2θ
]π2π/3
=
(π −
(2π
3−√3
2
))=
π
3+
√3
2
8. (16 points) Write the word TRUE (if always true) or FALSE. For this problem, no explanation isnecessary.
(a) If∞∑n=1
an diverges, then∞∑n=1
|an| diverges.
(b) The length of the curve r2 = 2 cos θ is given by L =
∫ π/2
04
√2 cos θ +
sin2 θ
2 cos θdθ.
(c) If the parametric equations x = f(t) and y = g(t) are twice differentiable, thend2y
dx2=d2y/dt2
d2x/dt2.
(d) The second order Taylor polynomial T2(x) centered at a = 1 is used to approximate f(x) =√x
on the interval [1, 1.1]. An estimate for the error of the approximation using Taylor’s Formula is1
16(0.1)3.
Solution:
(a) True because if∞∑n=1
|an| converges, then so does∞∑n=1
an.
(b) True. Note that r is not defined for cos θ < 0. If r =√2 cos θ, then
dr
dθ=− sin θ√2 cos θ
. The length
of the entire curve is four times the length for θ = 0 to π/2.
L =
∫ √r2 +
(dr
dθ
)2
dθ = 4
∫ π/2
0
√2 cos θ +
sin2 θ
2 cos θdθ.
(c) False.d2y
dx2=
ddt(dy/dx)
dx/dt
(d) True. R2(x) =f ′′′(z)
3!(x − 1)3 for z between x and 1. On the interval [1, 1.1], |f ′′′(z)| =∣∣∣ 3
8z5/2
∣∣∣ ≤ 38 and
∣∣(x− 1)3∣∣ ≤ (0.1)3, so |R2(x)| ≤ 3
8 ·13! · (0.1)
3 = 116(0.1)
3.
9. (12 points) Match the graphs shown below to the following equations. No explanation is necessary.
(a) x2 = 1 + y2/2 (d) r2 = 4/θ(b) x2 = 1− y2/4 (e) r2 = (cos θ)/4(c) x = 2 cos t, y = sin t (f) r = 1/2 + cos(2θ)
-2 2x
-2
2
y
H1L
-2 2x
-2
2
y
H2L
-2 2x
-2
2
y
H3L
-2 2x
-2
2
y
H4L
Solution: (1) d (2) f (3) c (4) a