Problem Points Score
M161, Final, Spring 2007
Name:
Section:
Instructor:
Time: 110 minutes. You may not use calculators on this
exam
d
sin(x) = cos(x),
dx
d
csc(x) = − csc(x) cot(x),
dx
d
sinh(x) = cosh(x),
dx
d
1
asin(x) = √
,
dx
1 − x2
d
1
,
acsc(x) = − √ 2
dx
x x −1
1
d
asinh(x) = √
,
dx
1 + x2
sin(2x) = 2 sin(x) cos(x)
tan2 (x) + 1 = sec2 (x)
d
cos(x) = − sin(x),
dx
d
sec(x) = sec(x) tan(x),
dx
d
cosh(x) = sinh(x),
dx
d
1
acos(x) = − √
,
dx
1 − x2
d
1
,
asec(x) = √ 2
dx
x x −1
1
d
acosh(x) = √ 2
,
x −1
Zdx
ln xdx = x ln x − x + C
1 + cos(2x)
cos2 (x) =
2
1
30
2ab
30
2cd
30
3
30
4
25
5
25
6
30
X
200
d
tan(x) = sec2 (x),
dx
d
cot(x) = − csc2 (x),
dx
d
tanh(x) = sech2 (x),
dx
d
1
atan(x) =
,
dx
1 + x2
d
1
acot(x) = −
dx
1 + x2
d
1
atanh(x) =
,
1 − x2
Zdx
sec(x)dx = ln(sec(x) + tan(x)) + C
1 − cos(2x)
sin2 (x) =
2
Taylor series of the function f (x) about x = a:
∞
X
f 0 (a)
f 00 (a)
f (n) (a)
2
f (a) +
(x − a) +
(x − a) + · · · =
(x − a)n
1!
2!
n!
n=0
Theorem (The Derivative Rule for Inverses) If f has an interval I as domain and f 0 (x)
exists and is never zero on I, then f −1 is differentiable at every point in its domain. The
1
value of (f −1 )0 at a point b = f (a) in the domain of f −1 is given by (f −1 )0 (b) = 0 .
f (a)
Length of a polar Curve: L =
Z
β
α
v
u
u
t
dr
r2 +
dθ
!2
dθ
1) a) Sketch the curve r = 2 + 2 sin(θ) for 0 ≤ θ ≤ 2π in the following diagram:
5
4
3
2
1
0
−5
−4
−3
−2
−1
−1
0
1
2
3
−2
−3
−4
−5
b) Calculate the area enclosed by the curve r = 2 + 2 sin(θ).
4
5
2) Evaluate the following integrals. Show your work (I.e. if an integral diverges, give a
proof).
Z
1
dx
a)
2
x − 4x + 5
√
b)
Z
e x
√ dx
x
c)
d)
Z
√
Z
1
2x2
dx (You may assume that |x| ≤ 3.)
9 − x2
∞
x3
1
dx
+x
∞
X
(x + 1)2n
converges. (I.e. deter9n
mine the radius of convergence and check both endpoints. Credit will be split evenly for
interval and each endpoint.)
3) Determine the interval on which the power series
n=0
4) We want to perform a partial fraction decomposition of the quotient
1
2
(x + 2) (x − 2) (x2 + 4)2 (x2 + x + 1)
Circle below which of the listed terms will (for suitable values of the coefficients) occur in this
decomposition (there are 6 terms occuring in total. You do not need to solve for the values
of the variables, perform an actual partial fraction decomposition or find an antiderivative).
You get 4 points for every correct term, -2 points for every wrong term. You cannot get less
than 0 points in this problem.
A) Ax + B
B)
A
Bx
C)
A
x−1
D)
E)
A
x+2
F)
Ax + B
x+2
G)
A
(x + 2)2
H)
Ax + B
(x + 2)2
I)
A
(x − 2)
J)
Ax + B
(x − 2)
K)
A
(x − 2)2
L)
Ax + B
(x − 2)2
A
(x + 2)2
N)
Ax + B
(x + 2)2
O)
A
(x + 2)3
P)
Ax + B
(x + 2)3
A
+ 4)
R)
Ax + B
(x2 + 4)
S)
A
+ 4)2
T)
Ax + B
(x2 + 4)2
A
(x2 + x + 1)
V)
Ax + B
(x2 + x + 1)
W)
X)
Ax + B
(x2 + x + 1)2
M)
Q)
U)
(x2
(x2
A
(x2 + x + 1)2
x5
A
+x+7
5) Match the Taylor series with the functions. Place the letter of the appropriate
series in the left column.
4
4 − x2
sin(x +
e−x
π
)
2
2
8
8 + x2
A
∞
X
(−1)n
1 2n
x
(2n)!
(−1)n
1 2n
x
23n
n=0
B
∞
X
n=0
C
D
∞
X
1
n
n=0 2
∞
X
n=0
x2n
(−1)n
1 2n
x
n!
6) a) Complete the following definition for limit of a sequence:
The sequence {an } converges to the number L if for every positive ε there exists an integer
N such that
12n − 1
and let L = 2. For a given ε > 0 calculate
6n + 1
N such that the condition in a) is fulfilled.
b) Now consider the sequence an =
c) Using the definition of convergence from a) write a proof that lim an = 2. (No points
n→∞
will be given for methods that are not N — proofs!)