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I have solved exam papers of Calculus. This is one of them, you can find all in my posts. Enjoy students. Some points of this solved exam paper are: Diverges by Comparison, Improper, Converges, Evaluate, Exists, Improper Integral, Function, Satisfies, Taylor Polynomial, Fourth Degree
Typology: Exams
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EXAM II - MARCH 10, 2006
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
2 EXAM II - MARCH 10, 2006
1
(1 + √x)^3 dx
For x ≥ 1 , (1 + √x)^3 > (√x)^3 = x^3 /^2. Thus, 0 <
1
(1 + √x)^3 dx <
1
x−^3 /^2 dx
= lim b→∞
∫ (^) b 1
x−^3 /^2 dx
= lim b→∞ − 2 x−^1 /^2
b 1
Hence, the improper integral converges.
(10 pts.)(b) (^) ∫ ∞ 2
cos^2 x x^2 dx.
Since 0 ≤ cos^2 x ≤ 1 , we have 0 < cos x^22 x≤ (^) x^12 when x ≥ 2 , and ∫ (^) ∞ 2
cos^2 x x^2 dx^ ≤
2
x^2 dx <
1
x^2 dx, which converges. It follows that (^) ∫ (^) ∞
2
cos^2 x x^2 dx converges as well.
4 EXAM II - MARCH 10, 2006
(8 pts.)(a) Write down the fourth-degree Taylor polynomial P 4 (x) for f near x = 2. The required polynomial is given by P 4 (x) = (−1) + 0 · (x − 2) + (− 2!2) (x − 2)^2 + (3) 3! (x − 2)^3 + (− 4!4) (x − 2)^4
= − 1 − (x 2 )^2 +^12 (x − 2)^3 − 16 (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 M 3 (x) = g(0) + g′(0)x + g
2! x
(^2) + g′′′(0) 3! x
3
= − 1 − x^2 +^12 x^3.
(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?
c h(x)
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. 5 c + c + c) = 2. 5 c. It follows that 2. 5 c = 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 M 3 (x) for f.
First we need to find f (n)(0). Note that f ′(x) = (^) 1+^1 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,
M 3 (x) = 0 + x + − 2!^1 x^2 + 3!^2 x^3
= x − x
2 2 +^
x^3
(8 pts.)(b) What is the maximum possible error committed by using M 3 (x) to estimate f (x) = ln(1 + x) over the interval [0, 1]?
The maximum error committed is no greater than or equal to K 4 4! |x|^4. For^0 ≤^ x^ ≤^1 , we have^ |x^4 | ≤^1 and^ K^4 = max^ |f^ (4)(x)|. From part (a), we have f (4)(x) = −6(1 + x)−^4 = (^) (1+−^6 x) 4. Thus, we let K 4 = 6. In other words, the maximum error committed by using M 3 (x) is no greater than (^) 4!^6 = 14.
