math 152 calculus ii name: spring 2019 sample final 1...

MATH 152 Calculus II Name:Spring 2019Sample Final 1 Signature:

Time Limit: 120 Minutes Instructor’s name

Grade Table(for instructor use only)

Question Points Score

1 5

2 5

3 5

4 5

5 5

6 5

7 5

8 5

9 5

10 5

11 5

12 5

13 5

Total: 65

1. This exam contains 12 pages (including this coverpage) and 13 questions.

2. Total of points of this exam is 65.

3. This is a closed book and notes exam.

4. Only non-graphing calculator is allowed.

5. You can do your preliminary work on scratch paper,but you must write your solutions on the exam paperin the space provided. Solutions on scratch paper willnot be graded.

6. Show and explain work for full credit.

7. By signing your name, you agree that all work is yourown and uninfluenced by others’, and you have com-pleted this exam in accordance with the instructionsand Illinois Institute of Technology’s Code of Aca-demic Honesty as outlined in the Student Handbook.

MATH 152 Calculus II Sample Final 1 - of 12

1. (5 points) Evaluate the indefinite integral∫cos−1 x dx

Solution: Recall that

d

dx

(cos−1 x

)= − 1√

1− x2, for x ∈ (−1, 1),

hence from integration by parts, we have∫cos−1x = x cos−1 x−

∫x

(−1√

1− x2

)dx

= x cos−1 x− 1

2

∫ (−2x√1− x2

)dx

= x cos−1 x− 1

2

(2√

1− x2)

+ C, u-substitution, with u = 1− x2

= x cos−1 x−(√

1− x2)

+ C, where C is a constant.

Note: With u = 1− x2, we have, du = −2x dx, and substitution technique gives∫−2x√1− x2

dx =

∫du√u

= 2√u+ C = 2

√1− x2 + C.

2. (5 points) Find the three cube roots of 8, and write your answer in the form of a + ib,where a, b are real numbers.

Solution: From De Moivre’s Theorem, the three cube roots of 8 are given by

wk = 813

[cos

(0 + 2kπ

3

)+ i sin

(0 + 2kπ

3

)], k = 0, 1, 2.

Thus, the three cube roots of 8 are

w0 = 2 (cos 0 + i sin 0) = 2 + i0

w1 = 2

(cos

2π

3+ i sin

2π

3

)= −1 + i

√3

w2 = 2

(cos

4π

3+ i sin

4π

3

)= −1− i

√3


3. (5 points) Evaluate the indefinite integral∫tan4 x sec6 x dx.

Solution: Let u = tanx, hence du = sec2 x dx. Now, using the trigonometric identitysec2 x = 1 + tan2 x, we have∫

tan4 x sec6 x dx =

∫tan4 x sec4 x(sec2 x) dx

=

∫tan4 x(1 + tan2 x)2(sec2 x) dx

=

∫u4(1 + u2)2 du

=

∫u8 + 2u6 + u4 du

=1

9u9 +

2

7u7 +

1

5u5 + C

=1

9tan9 x+

2

7tan7 x+

1

5tan5 x+ C, where C is a constant.

4. (5 points) Find the following limit

limx→0+

(5x+ 1)cotx

Solution: Since limx→0+ cotx = +∞, hence we have the case of indeterminate power. Lety = (5x+ 1)cotx, and taking the natural logarithm, we have

ln y = ln(5x+ 1)cotx

= cotx ln(5x+ 1)

So

limx→0+

ln y = limx→0+

ln(5x+ 1)

tanx

H= lim

x→0+

55x+1

sec2 x= 5

This implies thatlimx→0+

(5x+ 1)cotx = limx→0+

eln y

= elimx→0+ ln y = e5


5. (5 points) Evaluate the integral∫x3 − 2x2 + 2x− 5

x4 + 4x2 + 3dx.

Solution: First, we will decompose x3−2x2+2x−5x4+4x2+3

into partial fraction.

x3 − 2x2 + 2x− 5

x4 + 4x2 + 3=x3 − 2x2 + 2x− 5

(x2 + 1)(x2 + 3)=Ax+B

x2 + 1+Cx+D

x2 + 3.

Multiplying both sides by (x2 + 1)(x2 + 3) to get

x3 − 2x2 + 2x− 5 = (Ax+B)(x2 + 3) + (Cx+D)(x2 + 1)

= Ax3 +Bx2 + 3Ax+ 3B + Cx3 +Dx2 + Cx+D

= (A+ C)x3 + (B +D)x2 + (3A+ C)x+ (3B +D)

Comparing coefficients gives us the following system of equations:

A+ C = 1 (1)

B +D = −2 (2)

3A+ C = 2 (3)

3B +D = −5 (4)

Subtracting equation (1) from equation (3) gives us 2A = 1, that is A = 12 , and hence C = 1

2 .

Subtracting equation (2) from equation (4) gives us 2B = −3, that is B = −32 , and hence

D = −12 .

Therefore, we have

x3 − 2x2 + 2x− 5

x4 + 4x2 + 3=x3 − 2x2 + 2x− 5

(x2 + 1)(x2 + 3)=

12x−

32

x2 + 1+

12x−

12

x2 + 3.

∫x3 − 2x2 + 2x− 5

x4 + 4x2 + 3dx =

∫ ( 12x−

32

x2 + 1+

12x−

12

x2 + 3

)dx

=

∫ ( 12x

x2 + 1−

32

x2 + 1+

12x

x2 + 3−

12

x2 + 3

)dx

=1

4ln(x2 + 1)− 3

2tan−1 x+

1

4ln(x2 + 3)− 1

2√

3tan−1

(x√3

)+ C,

where C is a constant.


6. (5 points) Given the sequence {cos(n2π)

n2π

}∞n=1

(a) Determine the sequence is increasing, decreasing, or non-monotonic.

(b) Determine the sequence is bounded. If so, state a lower and an upper bound of thissequence.

(c) Determine if the sequence is convergent or divergent. If convergent, find its limit.

Solution:

(a) The sequence is non-monotonic, as the cosine function is oscillatory.

(b) Because ∣∣∣∣cos(n2π)

n2π

∣∣∣∣ ≤ | cos(n2π)|n2π

≤ 1

n2π≤ 1

π,

we see that the sequence is bounded, and we can choose − 1π as a lower bound, and 1

π asan upper bound of the sequence.

(c) Also since ∣∣∣∣cos(n2π)

n2π

∣∣∣∣ ≤ 1

n2π

we have

limn→∞

∣∣∣∣cos(n2π)

n2π

∣∣∣∣ ≤ limn→∞

1

n2π= 0,

which implies that

limn→∞

cos(n2π)

n2π= 0.


7. (5 points) Determine if the following series

∞∑n=1

(1 +

7

n

)2

e−n

is convergent or divergent. Give complete reasons for your answer.

Solution: We can use the Limit Comparison Test with an =(1 + 7

n

)2e−n and bn = e−n.

Since the limit

limn→∞

anbn

= limn→∞

(1 +

7

n

)2

= 1 > 0,

and that∞∑n=1

e−n =∞∑n=1

1

en

is a convergent geometric series, with common ration r = 1e < 1, by using Limit Comparison

Test, we conclude that the series∞∑n=1

(1 +

7

n

)2

e−n

also converges.

Comparison Test can also be used in this case, for example, notice that(1 +

7

n

)2

e−n ≤ (1 + 7)2 e−n ≤ 64e−n,

hence∞∑n=1

(1 +

7

n

)2

e−n

converges by Comparison Test, as∞∑n=1

64

en

is a convergent geometric series.


8. (5 points) Determine whether the given series

∞∑n=1

(−1)n+1 n2

n3 + 4

is absolutely or conditionally convergent, or if it diverges. Give complete reasons foryour answer.

Solution: Notice that∞∑n=1

∣∣∣∣(−1)n+1 n2

n3 + 4

∣∣∣∣ =

∞∑n=1

n2

n3 + 4.

Sincen2

n3 + 4≥ n2

n3 + 4n3=

n2

5n3≥ 1

5n,

and that∑∞

n=11n diverges (p-series with p = 1), the series

∞∑n=1

n2

n3 + 4

diverges by the Comparison Test.

Let an = n2

n3+4, we see that {an} is decreasing for n ≥ 2 since

d

dx

(x2

x3 + 4

)=

(x3 + 4)(2x)− x2(3x2)(x3 + 4)2

=x(8− x3)(x3 + 4)2

< 0, whenever x > 2.

Also,

limn→∞

an = limn→∞

1/n

1 + 4/n3= 0.

Thus, the series∞∑n=1

(−1)n+1 n2

n3 + 4

converges by the Alternating Series Test. Therefore

∞∑n=1

(−1)n+1 n2

n3 + 4

converges conditionally, but not absolutely.


9. (5 points) Find the center and radius of convergence of the series

∞∑n=1

(2x− 1)n

5n√n

.

Give complete reasons for your answer.

Solution: Let

an =(2x− 1)n

5n√n

,

then

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣ (2x− 1)n+1

5n+1√n+ 1

· 5n√n

(2x− 1)n

∣∣∣∣= lim

n→∞

|2x− 1|5

√n

n+ 1

= limn→∞

|2x− 1|5

√1

1 + 1/n

=|2x− 1|

5.

By the Ratio Test, the series∑∞

n=1(2x−1)n5n√n

converges when |2x−1|5 < 1, that is, |2x − 1| < 5

which is equivalent to

|x− 1

2| < 5

2.

From here we see that the center of the series is 12 , and the radius of convergence is R = 5

2 .


10. (5 points)

(a) Use your knowledge of geometric series or otherwise, find the Maclaurin series forthe function

f(x) =1

1 + x2.

Express your answer using the sigma notation.

(b) Show that

tan−1 x =∞∑n=0

(−1)nx2n+1

2n+ 1, for |x| < 1.

Solution:

(a) We know that∞∑n=0

xn =1

1− x, for |x| < 1,

now, replacing x by −x2, we have

1

1 + x2=

1

1− (−x2)=

∞∑n=0

(−x2)n =

∞∑n=0

(−1)nx2n.

Because this is a geometric series, it converges when | − x2| < 1, that is, x2 < 1, orequivalently, |x| < 1. Therefore the interval of convergence is still (−1, 1).

(b) Using the theorem (see Text, Page 794) which says that we can integrate term-by-termwithout affecting the radius of convergence, we have

tan−1 x =

∫1

1 + x2dx

=

∫ (1− x2 + x4 − x6 + · · ·

)dx

= x− 1

3x3 +

1

5x5 − 1

7x7 + · · ·+ C,

where x ∈ (−1, 1) and C is a constant. To find the value of C, we put x = 0 and obtainC = tan−1 0 = 0. Therefore we have

tan−1 x = x− 1

3x3 +

1

5x5 − 1

7x7 + · · ·

=

∞∑n=0

(−1)nx2n+1

2n+ 1,

for |x| < 1.

Note: The series actually also converges at x = −1 and x = 1, that is, the interval ofconvergence for the above series is [−1, 1], but you are not asked to show that.


11. (5 points)-0.5 0.5

x

-0.5

0.5

The figure on the left shows the polar curvewith equation

r = sin 4θ.

Find the area of the shaded region.

Solution: Notice that the region enclosed is swept out by a ray that rotates from 0 to π4 .

(This can be seen by setting r = 0, and obtained 4θ = nπ, where n ∈ Z.)

Hence the area of the enclosed region is given by

A =

∫ π4

0

1

2(sin 4θ)2 dθ

=1

2

∫ π4

0sin2 4θ dθ

=1

2

∫ π4

0

1

2(1− cos 8θ) dθ

=1

4

[θ − 1

8sin 8θ

]π4

0

=1

4

(π4

)=

1

16π


12. (5 points) Match the differential equation with its direction field. No justification needed.

-10 -5 5 10

-10

-5

5

(a)

-10 -5 5 10

-10

-5

5

10

(b)

-10 -5 5 10

-10

-5

5

10

(c)

-10 -5 5 10

-10

-5

5

(d)

dydx

= e−0.01xy2

Enter (a)-(d) here

dydx

= sinx cos y Enter (a)-(d) here

dydx

= 1− xy Enter (a)-(d) here

dydx

= xy

Enter (a)-(d) here

Solution: (b) (d) (c) (a)


13. (5 points) Solve the differential equation

(x+ 2)2dy

dx= 5− 8y − 4xy.

Solution: First we rewrite the differential equation into its standard form

dy

dx+ P (x)y = Q(x),

that isdy

dx+

4

x+ 2y =

5

(x+ 2)2. (∗)

Hence the integrating factor is

e∫

4x+2

dx = e4 ln |x+2| = (x+ 2)4.

Multiplying Equation (∗) above with the integrating factor, we obtained

(x+ 2)4dy

dx+ 4(x+ 2)3y = 5(x+ 2)2,

that is,d

dx

[(x+ 2)4y

]= 5(x+ 2)2 (∗∗)

Integrating Equation (∗∗) with respect to x we obtained

y =5

3(x+ 2)−1 + C(x+ 2)−4,

where C is a constant.

math 152 calculus ii name: spring 2019 sample final 1...

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