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**MATH106B CALCULUS II - PROF. P. WONG
**

EXAM II - MARCH 10, 2006

**NAME:
**

Instruction: Read each question carefully. Explain **ALL **your work and

give reasons to support your answers.

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

**Problems Maximum Score Your Score
**

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

or diverges by comparison. Justify your answers.

(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.

Justify your answers.

(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.

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