PRACTICE FINAL EXAM
Math 1220 sec 006
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by writing my name I swear by the honor code
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Read all of the following information before starting the exam:
• Show all work, clearly and in order, if you want to get full credit. I reserve the right to
take off points if I cannot see how you arrived at your answer (even if your final answer is
correct).
• No calculators, notes, or books are allowed for this test.
• Turn off cell phones, ipods, etc.
• No headphones allowed, turn hats backward, keep your eyes on your own paper.
• Please have your ID out and available.
• Please keep your written answers brief; be clear and to the point. I will take points off for
rambling and for incorrect or irrelevant statements.
• You have 120 minutes to work on this exam.
• Good luck!
Problem (value)
1 (10 points)
2 (10 points)
3 (10 points)
4 (10 points)
5 (10 points)
6 (15 points)
7 (10 points)
8 (10 points)
Total (85 points)
Score
1.
(10 points) Short Answer. No work required.
a. (2 pts) Complete the definition. A sequence of real numbers an is said to converge to
L if and only if for all > 0 ...
b. (2 pts)
The time for an exponentially decaying quantity y to go from size y0 to size
y0 /2 is called what?
c. (2 pts)
lim (1 + h)1/h =
h→0
d. (2 pts)
Hyperbolic sine is defined by sinh(x) =
e. (2 pts)
R∞
1
1/(xp )dx converges if and only if p is
2.
(10 points) Short Answer. No work required.
a. (2 pts)
The integral
b. (2 pts)
A series
∞
P
R1
0
√1 dx
x
does not exist in the proper sense because
an is said to converge if and only if
n=1
c. (2 pts)
A series of the form a0 + a1 x + a2 x2 + ... is called a
d. (2 pts) If I perform a Taylor series expansion of function f (x) to the nth order about
x = a then the remainder term is represented as Rn (x) =
e. (2 pts)
Tangent lines to the curve r = f (θ) at the origin can be found by solving
3.
(10 points)
a. (5 pts)
Calculate the integrating factor in the first order linear differential equation
x2
b. (5 pts)
Evaluate
R2
1
ln(x)dx
dy
= −3xy + 5x5
dx
4.
(10 points)
a. (5 pts)
Find the partial fractions decomposition of
b. (5 pts)
Evaluate, if possible, the integral
R2
0
√ dx .
4−x2
x
.
(x−3)2
5.
(10 points)
a. (5 pts)
x = 0.
b. (5 pts)
Find the Taylor polynomial of order 4 of f (x) = ln(1 + x) centered about
Prove that ln xr = r ln x.
6.
(15 points) Determine if the following series converge or diverge and give justification for
your conclusion.
∞
P
2
(−1)n+1 nen
a. (5 pts)
n=1
b. (5 pts)
∞
P
n=1
c. (5 pts)
ln n
n
Find the convergence set for the series
∞
P
n=0
n!xn .
7.
(10 points)
a. (2 pts)
Find the Cartesian coordinates of the point with polar coordinates (−1, π/2)
b. (3 pts)
Transform the equation x2 = 4py into polar coordinates.
c. (5 pts)
closed curve.)
Find the area inside the limacon r = 2 + cos θ. (Note: a limacon is a simple
8.
(10 points)
a. (5 pts)
Consider the following statement and attempt at a proof.
Claim: if lim nan = 1, then
n→∞
∞
P
an diverges.
n=1
Proof: Let lim nan = 1. We must show that
n→∞
∞
P
an diverges. If lim nan = 1 then there is some
n=1
n→∞
N such that an ≥ 0 for all n > N because otherwise the limit couldn’t be positive. Let bn =
∞
P
an diverges.
so lim abnn = lim nan = 1. Since this limit is not equal to zero the series
n→∞
n→∞
n=1
What is wrong with this proof?
b. (5 pts)
If possible, find the limit
lim (tan x)cos x
x→π/2−
1
n
Appendix: Tables of useful integrals and series convergence tests
Trig identities
sin(−x) = − sin x
cos(−x) = cos x
tan(−x) = − tan x
sin
cos
tan
π
2 − x = cos x
π
2 − x = sin x
π
2 − x = cot x
sin2 x + cos2 x = 1
1 + tan2 x = sec2
1 + cot2 x = csc2
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x
= 2 cos2 x − 1
= 1 − 2 sin2 x
sin x sin y = − 12 [cos(x + y) − cos(x − y)]
cos x cos y = 12 [cos(x + y) + cos(x − y)]
sin x cos y = 12 [sin(x + y) + sin(x − y)]
Integrals
R
tan udu = − ln | cos u| + C
R
sec udu = ln | sec u + tan u| + C
R
√ du
a2 −u2
R
du
a2 +u2
=
1
a
R
du
a2 −u2
=
1
2a
ln u+a
u−a + C
R
√ du
u u2 −a2
=
1
a
R
sin2 udu = 12 u − 14 sin 2u + C
R
cos2 udu = 21 u + 14 sin 2u + C
R
tan2 udu = tan u − u + C
= sin−1
u
a
tan−1
+C
u
a
+C
sec−1 ua + C
q
x
sin(x/2) = ± 1−cos
q 2
x
cos(x/2) = ± 1+cos
2