# Diverges by Comparison - Calculus - Solved Exam, Exams for Calculus. Institute of Mathematics and Applications

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Exam2-soln.DVI

MATH106B CALCULUS II - PROF. P. WONG

EXAM II - MARCH 10, 2006

NAME:

Advice: DON’T spend too much time on a single problem.

1. 20

2. 20

3. 20

4. 18

5. 22

Total 100

1

2 EXAM II - MARCH 10, 2006

1. Determine whether each of the following improper integrals converges

(10 pts.)(a) ∫ ∞

1

1 (1 +

√ x)3

dx

For x ≥ 1, (1+ √

x)3 > ( √

x)3 = x3/2. Thus,

0 < ∫ ∞

1

1 (1 +

√ x)3

dx <

∫ ∞

1 x−3/2 dx

= lim b→∞

∫ b

1 x−3/2 dx

= lim b→∞

−2x−1/2 ∣∣∣∣ b

1

= 2 < ∞.

Hence, the improper integral converges.

(10 pts.)(b) ∫ ∞

2

cos2 x x2

dx.

Since 0 ≤ cos2 x ≤ 1, we have 0 < cos2 x x2

≤ 1 x2

when x ≥ 2, and ∫ ∞

2

cos2 x x2

dx ≤ ∫ ∞

2

1 x2

dx <

∫ ∞

1

1 x2

dx,

which converges.

It follows that ∫ ∞

2

cos2 x x2

dx

converges as well.

MATH106B CALCULUS II - PROF. P. WONG 3

2. For each of the following improper integrals, evaluate if it exists.

(10 pts.)(a) ∫ ∞

e

1 x lnx

dx.

Let u = lnx so du = 1xdx. When x = e, u = 1 and when x = b, u =

ln b. Thus, ∫ ∞

e

1 x lnx

dx = lim b→∞

∫ b

e

1 x lnx

dx

= lim b→∞

∫ ln b

1

1 u

du = lim b→∞

ln |u| ∣∣∣∣ ln b

1

= lim b→∞

ln | lnb| − 0 = +∞.

Hence, the improper integral diverges.

(10 pts.)(b) ∫ 2

−1

1 3 √

x dx

[Hint: this is an improper integral]

The integral is improper at x = 0. Thus, we write ∫ 2

−1

1 3 √

x dx =

∫ 0

−1 x−1/3dx+

∫ 2

0

x−1/3dx

= lim b→0

∫ b

−1 x−1/3dx + lim

c→0

∫ 2

c x−1/3dx

= lim b→0

x2/3

2/3

∣∣∣∣ b

−1 + lim

c→0

x2/3

2/3

∣∣∣∣ 2

c

= lim b→0

3 2 (b2/3 − 1) + lim

c→0

3 2 (22/3 − c2/3) = 3

2 (22/3 − 1).

4 EXAM II - MARCH 10, 2006

3. Suppose a function f satisfies

f(2) = −1, f ′(2) = 0, f ′′(2) = −2, f ′′′(2) = 3, f (4)(2) = −4.

(8 pts.)(a) Write down the fourth-degree Taylor polynomial P4(x) for f near

x = 2.

The required polynomial is given by

P4(x) = (−1) + 0 · (x − 2) + (−2) 2!

(x − 2)2 + (3) 3!

(x − 2)3 + (−4) 4!

(x− 2)4

= −1 − (x2)2 + 1 2 (x − 2)3 − 1

6 (x− 2)4.

(5 pts.)(b) If g(x) = f(x+2), find the third-degree Maclaurin polynomial

for g(x). [Hint: Use part (a).]

Note that g(n)(0) = f (n)(2). It follows that

M3(x) = g(0) + g′(0)x+ g′′(0) 2!

x2 + g′′′(0) 3!

x3

= −1− x2 + 1 2 x3.

(7 pts.)(c) The graph of a function h(x) is given below. For what value

of c is h(x) a probability density function?

0 1 4

c h(x)

2

For the function h(x) to be a probability density function, we

need to check that h(x) ≥ 0 and ∫∞ −∞ h(x) dx = 1. The first condition

is satisfied. For the second condition, the improper integral gives

the area under the graph of h(x) which is (0.5c + c + c) = 2.5c. It

follows that 2.5c = 1 or c = 25.

MATH106B CALCULUS II - PROF. P. WONG 5

4.(10 pts.)(a) Let f(x) = ln(1 + x). Find the third-degree Maclaurin

polynomial M3(x) for f .

First we need to find f (n)(0). Note that f ′(x) = 11+x = (1 + x) −1,

f ′′(x) = −(1 + x)−2, and f ′′′(x) = 2(1 + x)−3. It follows that f(0) = ln 1 = 0, f ′(0) = 1, f ′′(0) = −1, and f ′′′(0) = 2. Thus,

M3(x) = 0 + x + −1 2!

x2 + 2 3!

x3

= x − x 2

2 +

x3

3 .

(8 pts.)(b) What is the maximum possible error committed by using

M3(x) to estimate f(x) = ln(1 + x) over the interval [0, 1]?

The maximum error committed is no greater than or equal to K4 4! |x|

4. For 0 ≤ x ≤ 1, we have |x4| ≤ 1 and K4 = max |f (4)(x)|. From part (a), we have f (4)(x) = −6(1+x)−4 = −6

(1+x)4 . Thus, we let K4 = 6.

In other words, the maximum error committed by using M3(x) is

no greater than 64! = 1 4 .

6 EXAM II - MARCH 10, 2006

5. Evaluate each of the following indefinite integral.

(11 pts.)(a) ∫ 2t√ 1− t4

dt.

Let u = t2 so that du = 2t dt. Therefore,

2t√ 1− t4

dt = ∫

du√ 1− u2

= arcsinu + C

= arcsin t2 + C.

(11 pts.)(b) ∫ 1

5 + 4x + x2 dx.

First, note that 5 + 4x + x2 = 1 + (4 + 4x + x2) = 1 + (x + 2)2.

Now let w = x + 2 so that dw = dx and∫ 1

5 + 4x + x2 dx =

∫ dw

1 + w2

= arctanw + C

= arctan(x + 2) + C.