M161, Test 2, Spring 2007
Name:
Section:
Instructor:
Time: 75 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
d
1
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
1
d
acos(x) = − √
,
dx
1 − x2
d
1
asec(x) = √ 2
,
dx
x x −1
d
1
,
acosh(x) = √ 2
x −1
Zdx
ln xdx = x ln x − x + C
1 + cos(2x)
cos2 (x) =
2
Problem Points Score
1
20
2ab
16
2cd
16
3
16
4
16
5
16+5
X
100+5
d
tan(x) = sec2 (x),
dx
d
cot(x) = − csc2 (x),
dx
d
tanh(x) = sech2 (x),
dx
1
d
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
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)
1) The following terms are prototypes of the terms resulting from a partial fraction decomposition.
In each case determine an antiderivative:
Z
1
dx
a)
(x − 1)
b)
c)
d)
Z
1
dx
(x − 1)2
Z
x+1
dx
(x2 + 1)
Z
(x2
x+1
dx
+ 2x + 2)
2) Z Evaluate the following integrals. Show your work.
sin5 x cos2 xdx
a)
b)
Z
x5 ln xdx
c)
Z
0
d)
Z
∞
(x2
√
x
dx
+ 4)3/2
x2
dx
4 − x2
3) For each of the following improper integrals explain whether the integral converges or
diverges.
(You do not need to calculate the values of convergent integrals.)
Z 1
1
dx
a)
3
0 x +x
b)
∞
1
dx
x3 + x
∞
1
dx
x2
∞
x
dx
ex
Z
1
c)
Z
0
d)
Z
1
4) We want to perform a partial fraction decomposition of the quotient
x11 − 4x10 + 8x9 − 17x8 + 4x7 + 56x6 − 112x5 + 256x4 − 320x3 + 128x2 − 255x + 259
(x + 2) (x − 2)3 (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 8 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 2 points for every correct term, -1 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)3
J)
Ax + B
(x − 2)3
K)
A
x+2
L)
Ax + B
x+2
x5
A
+x+7
M)
A
(x + 2)2
N)
Ax + B
(x + 2)2
O)
A
(x + 2)3
P)
Ax + B
(x + 2)3
Q)
A
(x2 + 4)
R)
Ax + B
(x2 + 4)
S)
A
(x2 + 4)2
T)
Ax + B
(x2 + 4)2
A
+ 4)3
V)
Ax + B
(x2 + 4)3
W)
A
(x2 + x + 1)2
Z)
Ax + B
(x2 + x + 1)2
U)
Y)
(x2
(x2
A
+ x + 1)
X)
Ax + B
+ x + 1)
(x2
5) In this problem we want to see that asinh(x) really is not a “new” function deserving
its own name.
Z
1
a) We know that asinh(x) + C = √ 2
dx. Integrate this using a suitable trigonometric
x +1
substitution to end up with a solution not involving hyperbolic functions.
b) Let F (x) be the antiderivative you obtained in Part a). We know that F (x)+C = asinh(x)
for some constant C. Determine the value of C.
c (bonus) ) Verify your calculation by showing that F (sinh(x)) + C = x.