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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.