COLUMBIA UNIVERSITY
Math V1102
Calculus II
Spring 2014
Practice Exam I
02.21.2014
Instructor: S. Ali Altug
Name and UNI:
Question:
1
2
3
4
5
6
Total
Points:
6
6
6
6
6
3
33
Score:
Instructions:
• There are 6 questions on this exam.
• Please write your NAME and UNI on top of EVERY page.
• SHOW YOUR WORK in every question.
• Please write neatly, and put your final answer in a box.
• No calculators, cell phones, books, notebooks, notes or cheat sheets are allowed.
02.21.2013
Final
Name & UNI:
1. (6 points)
Z
x ln(x)dx
Solution: We use integration by parts. Let u0 = ln(x) and v = x, then u = x ln(x) − x and v 0 = 1.
Let I denote the integral, i.e.
Z
I = x ln(x)dx
Z
x ln(x) = x2 ln(x) − x2 −
Z
(x ln(x) − x)dx
= x2 ln(x) − x2 − I +
x2
2
Hence we have,
1
2I = x2 ln(x) −
2
which implies that
x2
I=
2
1
ln(x) −
2
+C
2. (6 points)
Z
sec3 (x) tan3 (x)dx
Solution: We use the trigonometric identity sec2 (x) = tan2 +1 to express everything in terms of
sec(x) and it’s powers.
sec3 (x) tan3 (x) = sec2 (x) tan2 (x) sec(x) tan(x)
= sec2 (x)(sec2 (x) − 1) sec(x) tan(x)
= sec4 (x) sec(x) tan(x) − sec2 (x) sec(x) tan(x)
Substituting this identity in the integral gives
Z
Z
Z
sec3 (x) tan3 (x)dx = sec4 (x) sec(x) tan(x)dx − sec2 (x) sec(x) tan(x)dx
Now we use the substitution u = sec(x). Then du = sec(x) tan(x)dx and hence the integrals above
turn into,
Z
Z
Z
sec3 (x) tan3 (x)dx = sec4 (x) sec(x) tan(x)dx − sec2 (x) sec(x) tan(x)dx
Z
Z
4
= u du − u2 du
u5
u3
−
+C
5
3
5
sec (x) sec3 (x)
=
−
+C
5
3
=
3. (6 points)
Z
arcsin(x)dx
Page 2
02.21.2013
Final
Name & UNI:
(Hint: You can try substituting x = sin(θ), and then use integration by parts.)
Solution: We use the hint. Let x = sin(θ), then dx = cos(θ)dθ. Substituting this into our integral
gives,
Z
Z
arcsin(x)dx = θ cos(θ)dθ
We go on with the suggestion in the hint and integrate by parts. Let u0 = cos(θ) and v = θ, then
u = sin(θ) and v 0 = 1. This gives,
Z
Z
arcsin(x)dx = θ cos(θ)dθ
Z
= θ sin(θ) − sin(θ)dθ
= θ sin(θ) + cos(θ)
Now the only point remaining is to re-express everything in terms of theqvariable x. Recall that
√
x = sin(θ), then arcsin(x) = θ. Also since sin2 (θ) + cos2 (θ) = 1, cos(θ) = 1 − sin2 (θ) = 1 − x2 .
Substituting these back into the answer above, we get the final form of the answer:
Z
p
arcsin(x)dx = x arcsin(x) + 1 − x2 + C
4. Determine if the following improper integrals converge.
(a) (3 points)
π
Z
tan(x)dx
0
(b) (3 points)
∞
Z
√
1
(Hint: You can use the inequality
√
√
x−1<
dx
x−1
x.)
Solution:
(a) Note that tan(x) has an infinity type discontinuity at x =
Z
0
π
Z
tan(x)dx = lim−
t→ π2
π
2.
t
0
Therefore by definition,
Z
tan(x)dx + lim+
t→ π2
π
tan(x)dx
t
Let us first consider the first integral above.
Z t
t
tan(x)dx = lim ln | sec(x)||0
lim
−
t→ π2
−
t→ π2
0
= lim− (ln | sec(t)| − 0)
t→ π2
Rt
Since limt→ π− ln | sec(x)| = ∞, we conclude that limt→ π− 0 tan(x)dx does not exist, and hence
2
2
the integral diverges. (Note that we do not need to check the other part of the integral since
for the integral to converge both of the limits should exist.)
(b) We use the hint.
√
x−1<
√
x⇒ √
Page 3
1
1
>√
x−1
x
02.21.2013
Final
Therefore,
Z
∞
√
1
Name & UNI:
dx
>
x−1
Z
∞
1
dx
√
x
Since the latter integral diverges (by the p-test) we conclude that the original integral also
diverges.
5. (6 points)
Z
x5 + 1
dx
x4 − 1
Solution:
Since the degree of the numerator is greater than the degree of the denominator we first perform
long division. This gives,
x5 + 1 = x(x4 − 1) + x + 1
Hence,
x+1
x5 + 1
=x+ 4
x4 − 1
x −1
Now note that the denominator has the following factorization,
x4 − 1 = (x + 1)(x − 1)(x2 + 1)
Hence,
x+1
1
=
4
x −1
(x − 1)(x2 + 1)
Therefore by using particle fractions we can express the quotient
x+1
x4 −1
by
1
B1
A2 x + B2
=
+
(x − 1)(x2 − 1)
x−1
x2 + 1
We now solve for the coefficients A2 and B1 , B2 . Equating the denominators gives,
B1 (x2 + 1) + (A2 x + B2 )(x − 1)
1
=
2
(x − 1)(x − 1)
(x − 1)(x2 + 1)
(B1 + A2 )x2 + (B2 − A2 )x + B1 − B2
=
(x − 1)(x2 + 1)
Hence we get the following system of equations,
B1 + A2 = 0
B2 − A2 = 0
B1 − B2 = 1
Which has the solution B1 = 21 , A2 =
−1
2 , B2
=
−1
2 .
Hence,
Z 5
Z x +1
1 1
1 x+1
dx =
x+
−
dx
x4 − 1
2 x − 1 2 x2 + 1
p
x2
ln |x2 + 1| arctan(x)
=
+ ln |x − 1| −
−
+C
2
4
2
6. (3 points) Find the volume of the solid obtained by rotating the area between
around the x-axis.
Page 4
1
x
and
1
x2
from 1 to ∞
02.21.2013
Final
Solution: Each cross section has area π
1
x2
−
1
x4
Name & UNI:
, therefore the volume is
Z ∞
1
2π
1
−
dx =
π
2
4
x
x
15
1
Some Identities
•
sin2 (θ) + cos2 (θ) = 1
•
tan2 (θ) + 1 = sec2 (θ)
•
sin(2θ) = 2 sin(θ) cos(θ)
•
cos(2θ) = cos2 (θ) − sin2 (θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ)
Page 5