COLUMBIA UNIVERSITY
Math V1102
Calculus II
Fall2014
Midterm I
09.30.2014
Instructor: S. Ali Altuğ
B
Name and UNI:
Question:
1
2
3
4
5
6
7
Total
Points:
6
6
6
6
6
6
0
36
Score:
Instructions:
• There are 7 questions on this exam.
• Please write your NAME and UNI on top of EVERY page.
• In order to get full credit you need to answer the first 6 questions correctly.
• The last question is a bonus question, and you do not have to answer it.
• Unless otherwise is explicitly stated 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.
• Some useful 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)
09.30.2014
Midterm I
Name & UNI:
1. (6 points)
Z
3x2 − 2x + 1
dx
x3 − x2 + x
Solution: By partial fractions,
3x2 − 2x + 1
A
Bx + C
= + 2
3
2
x −x +x
x
x −x+1
Solving for A, B, C gives A = 1, B = 2, C = −1. Hence,
Z
Z
Z
3x2 − 2x + 1
1
2x − 1
dx
=
dx
+
dx
x3 − x2 + x
x
x2 − x + 1
= ln |x| + ln |x2 − x| + C
Where we used u = x2 − x + 1 substitution in the last integral.
Page 2
09.30.2014
Midterm I
Name & UNI:
2. (6 points)
Z
sin2 (x) cos4 (x)dx
Hint: Double angle formulas.
Solution:
By the double angle formulas we have sin2 (x) =
Z
2
Z 4
sin (x) cos (x)dx =
1−cos(2x)
2
1 − cos(2x)
2
and cos2 (x) =
1 + cos(2x)
2
Z
=
=
=
=
=
=
1+cos(2x)
.
2
Hence,
2
dx
1
1 − cos2 (2x) (1 + cos(2x)) dx
8
Z
1
sin2 (2x)(1 + cos(2x))dx
8
Z
Z
1
1
2
sin (x)dx +
sin2 (2x) cos(2x)dx
8
8
Z
Z
1
1
1 − cos(4x)
dx +
sin2 (2x) cos(2x)dx
8
2
8
Z
Z
Z
1
1
1
dx −
cos(4x)dx +
sin2 (2x) cos(2x)dx
16
16
8
x
sin(4x) sin3 (2x)
−
+
+C
16
64
48
Where we used u = sin(2x) substitution in the last integral.
Page 3
09.30.2014
Midterm I
Name & UNI:
3. (6 points)
π
2
Z
sin(6x) cos(2x)dx
0
Hint: Try integration by parts (most probably more than once).
Solution: Let u = sin(6x) and dv = cos(2x)dx, then integration by parts gives,
Z
Z
sin(6x) sin 2x)
sin(6x) cos(2x)dx =
− 3 cos(6x) sin(2x)dx
2
Then using integration by parts once again in the second integral with u = cos(6x) and dv =
sin(2x)dx gives
Z
Z
cos(6x) cos(2x)
cos(6x) sin(2x)dx = −
− 3 sin(6x) cos(2x)dx
2
Substituting this expression back into the first integration by parts then gives
Z
Z
sin(6x) sin(2x) + cos(6x) cos(2x)
+ 9 sin(6x) cos(2x)dx
sin(6x) cos(2x)dx =
2
R
Solving for sin(6x) cos(2x)dx then gives,
Z
sin(6x) sin(2x) + cos(6x) cos(2x)
sin(6x) cos(2x)dx = −
16
Finally substituting the boundaries we get,
Z
0
π
2
π
sin(6x) sin(2x) + cos(6x) cos(2x) 2
sin(6x) cos(2x)dx = −
16
0
=0
Page 4
09.30.2014
Midterm I
Name & UNI:
4. (6 points)
Z
x7
p
1 + x4 dx
Hint: Start with u = x2 substitution.
Solution:
Substituting u = x2 gives,
Z
Z
p
p
1
4
x 1 + x dx =
u3 1 + u2 du
2
7
We then use the substitution u = tan(θ) and get
Z
u3
p
1 + u2 du =
Z
Z
=
tan3 (θ) sec3 (θ)dθ
tan2 (θ) sec2 (θ) tan(θ) sec(θ)dθ
Z
(sec2 (θ) − 1) sec2 (θ) tan(θ) sec(θ)dθ
Z
Z
= sec4 (θ) tan(θ) sec(θ)dθ − sec2 (θ) tan(θ) sec(θ)dθ
=
=
sec5 (θ) sec3 (θ)
−
+C
5
3
Where we used u = sec(θ) substitution
variables
√ to evaluate the last integrals. We then substitute the √
back: Since u = tan(θ), sec(θ) = 1 + u2 . Then since u = x2 we get that sec(θ) = 1 + x4 .
Therefore the final answer is,
Z
5
x7
p
1 + x4 dx =
3
(1 + x4 ) 2
(1 + x4 ) 2
−
+C
10
6
Page 5
09.30.2014
Midterm I
Name & UNI:
5. (6 points)
Z
x+
1
√
dx
x2 − 1
Hint: Try x = 21 (eu + e−u ) substitution.
Solution:
We will use the hint. Let x = 21 (eu + e−u ) and without loss of generality assume u > 0 (the
substitution is symmetric with respect to u 7→ −u), then dx = 12 (eu − e−u ). Subbtituting these back
into the integral gives,
Z
Z
eu − e−u
1
√
q
dx =
x + x2 − 1
eu + e−u + 2 14 (e2u + 2 + e−2u ) − 1
Z
eu − e−u
q
=
eu + e−u + 2 14 (e2u − 2 + e−2u )
Z
eu − e−u
q
=
eu + e−u + 2 14 (eu − e−u )2
Z
eu − e−u
=
u
e + e−u + eu − e−u
Z u
1
e − e−u
=
du
2
eu
Z
Z
1
1
du −
e−2u du
=
2
2
u e−2u
= +
+C
2
4
Now in order to solve for x in terms of u we note the following: Let eu = y. Then y > 1 since
we chose u > 0, and the equation x = 12 (eu + e−u ) becomes 2x = y + y1 . This in particular
√
√
2
implies y 2 − 2xy + 1 = 0. Therefore y = 2x± 24x −4 = x ± x2 − 1. Since y > 1 we need to have
√
y = x + x2 − 1. Therefore we get,
p
eu = x + x2 − 1
Substituting this back into the answer then gives
√
Z
1
ln |x + x2 − 1|
1
√
√
dx =
+
+C
2
x + x2 − 1
4(x + x2 − 1)2
Page 6
09.30.2014
Midterm I
Name & UNI:
6. Determine if the following integrals converge or diverge.
(a) (3 points)
Z
∞
1
dx
+1
2x2
1
(b) (3 points)
1
Z
x ln(x)dx
0
Hint: For evaluating limits of the form limt→0+ x ln(x) you can use L’Hospital’s rule.
Solution:
(a) By definition,
Z
1
∞
1
dx = lim
2
t→∞
2x + 1
Z
t
1
dx
+1
Z1 ∞
1
√
dx
= lim
t→∞ 1
(x 2)2 + 1
√
Z t 2
1
1
= lim √ √
du
2+1
t→∞
u
2 2
√
1
t 2
= √ lim arctan(u)|√2
2 t→∞
√
√
1
= √ lim (arctan(t 2) − arctan( 2))
t→∞
2
√
π
2 − arctan( 2)
√
=
2
2x2
Therefore the integral is convergent.
(b) The integrand is continuous on (0, 1) and the only discontinuity is due to ln(x) and is at x = 0.
The integral is then defined by,
1
Z
0
Z
x ln(x)dx = lim+
t→0
1
x ln(x)
t
We calculate the latter integral by integration by parts. Let u = ln(x) and dv = xdx, then
Z
1
1
Z
x2 ln(x) 1 1
−
xdx
2
2 t
t
1
2x2 ln(x) − x2 =
4
t
x ln(x) =
t
1 2t2 ln(t) − t2
=− −
4
4
Substituting this back into the improper integral we get,
Z
1
1 2t2 ln(t) − t2
− −
4
4
t→0+
1 1
= − − lim+ t2 ln(t)
4 2 t→0
x ln(x)dx = lim
0
Page 7
09.30.2014
Midterm I
Name & UNI:
To calculate this last limit we use L’Hospital’s rule.
lim t2 ln(t) = lim
t→)+
t→)+
ln(t)
1/t2
= lim+ t2 ln(t)
t→)
=
−1
lim t2 ln(t)x2
2 t→)+
=0
Hence we get,
1
Z
1/t
−2/t3
x ln(x)dx = −
0
and in particular the integral converges.
Page 8
1
4
09.30.2014
Midterm I
Name & UNI:
7. (5 points (bonus)) Let n ≥ 0 be fixed. Calculate
Z
π
2
sinn (x)
dx
sin (x) + cosn (x)
n
0
Hint: Try coming up with a substitution that flips the boundary. This will result in an integral that
looks very similar to the original integral. You may then try to take a combination of the two forms of
the same integral. Hint-2: The result is independent of the specific choice of n.
Solution: The substitution that the hint is referring to is x = π2 − u. This transforms the integral
to
Z π2
sinn (x)
dx
I=
n
sin (x) + cosn (x)
0
Z π2
sinn (π/2 − u)
=
du
sinn (π/2 − u) + cosn (π/2 − u)
0
Z π2
cosn (u)
=
du
n
cos (u) + sinn (u)
0
Therefore we get,
Z
π
2
2I =
0
Z
=
sinn (x)
dx +
sinn (x) + cosn (x)
π
2
dx
0
=
Hence I =
π
2
π
4.
Page 9
Z
π
2
cosn (x)
dx
sin (x) + cosn (x)
n
0