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Calculus is most common subject I know so far. This is one of past exam papers you can find in my uploads. Key points of the exam are: Fourth Degree, Indefinite Integral, Evaluate, Fourth Degree, Maclaurin Polynomial, Third Degree, Taylor Polynomial, Maximum Possible Error, According, Committed
Typology: Exams
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EXAM II - NOVEMBER 4, 2011
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
Total 100
1
2 EXAM II - NOVEMBER 4, 2011
1.(10 pts.)(a) Evaluate the indefinite integral ∫ ln(1 + x^2 ) dx.
Let u = ln(1 + x^2 ) and dv = dx. Then
du = 2 x dx 1 + x^2 and v = x.
It follows that (^) ∫
ln(1 + x^2 ) dx IBP = ln(1 + x^2 ) · x −
x · 2 x dx 1 + x^2 = x ln(1 + x^2 ) −
2 x^2 1 + x^2
dx
= x ln(1 + x^2 ) −
1 + x^2
dx
= x ln(1 + x^2 ) − 2 x + 2 arctan x + C.
(10 pts.)(b) Evaluate the indefinite integral ∫ x + 5 x^2 + 3x − 4 dx.
Write x + 5 x^2 + 3x − 4
x + 4
x − 1
for some constants A and B. Thus,
x + 5 ≡ A(x − 1) + B(x + 4).
When x = 1, we have B = 65. When x = − 4 , we have A = −^15.
It follows that ∫ x + 5 x^2 + 3x − 4 dx =
x + 4 dx +
5 x − 1 dx
= −
ln |x + 4| +
ln |x − 1 | + C.
4 EXAM II - NOVEMBER 4, 2011
Since f (x) = ln(x + 1), we have
f ′(x) = (x + 1)−^1 , f ′′(x) = −(x + 1)−^2 , f ′′′(x) = 2(x + 1)−^3 , and f (4)(x) = −6(x + 1)−^4.
Thus, at x 0 = 0, we have f (0) = ln 1 = 0, f ′(0) = 1, f ′′(0) = − 1 , f ′′′(0) = 2, and f (4)(0) = − 6. Hence,
M 4 (x) = (0) + (1)x +
x^2 +
x^3 +
x^4
= x − x^2 2
x^3 3
x^4 4
(10 pts.)(b) Let g(x) = √x^1 +2. Find the third-degree Taylor polynomial P 3 (x) for g(x) based at x 0 = 2.
Since g(x) = (x + 2)−^1 /^2 , we have
g′(x) = −
(x + 2)−^3 /^2 , g′′(x) =
(x + 2)−^5 /^2 , and g′′′(x) = −
(x + 2)−^7 /^2.
Thus, at x 0 = 2, we have g(2) = 12 , g′(2) = − 161 , g′′(2) = 1283 , and g′′′(2) = − 102415. Hence,
P 3 (x) =
(x − 2) +
(x − 2)^2 −
(x − 2)^3
=
(x − 2) +
(x − 2)^2 −
(x − 2)^3.
MATH106B,C CALCULUS II - PROF. P. WONG 5
4.(10 pts.)(a) Let f (x) =
x. What is the maximum possible error, according to Taylor’s theorem, committed by using the third-degree Taylor polynomial P 3 (x) based at x 0 = 1 to estimate f (x) for 12 ≤ x ≤ 32?
According to Taylor’s theorem, we first need to find the fourth derivative of f. Since f (x) = x^1 /^2 , it follows that
f ′(x) =
x−^1 /^2 , f ′′(x) = −
x−^3 /^2 , f ′′′(x) =
x−^5 /^2 , and f (4)(x) = −
x−^7 /^2.
Over the interval 1 / 2 ≤ x ≤ 3 / 2 , |x−^7 /^2 | ≤ 27 /^2 so that we let
K 4 =
By Taylor’s theorem, the error committed by P 3 (x) is:
|f (x) − P 3 (x)| ≤
|x − 1 |^4 ≤
(10 pts.)(b) Let
h(x) =
kx^3 , if 0 ≤ x ≤ 2; 0 , elsewhere.
Here, k is a constant. Determine the value(s) of k for which h(x) is a probability density function. Justify your answer.
For h(x) to be a pdf, h(x) ≥ 0 for all x and
−∞ h(x)^ dx^ = 1. Since^ h(x) = 0^ outside of the interval [0, 2], it follows that ∫ (^) ∞
−∞
h(x) dx =
0
kx^3 dx =
kx^4 4
2
0
= 4k.
By setting 4 k = 1, we have k = 14. With k = 14 , h(x) is a pdf.
