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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: Value, Definite Integral, Indefinite Integral, Function, Satisfies, Third Degree, Taylor Polynomial, Possible Error, Committed, Maclaurin Polynomial
Typology: Exams
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EXAM II - MARCH 9, 2007
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 9, 2007
1.(10 pts.)(a) Find the exact value of the definite integral ∫ (^2) 1
x ln(x^2 ) dx.
Let u = ln(x^2 ) and dv = xdx. It follows that du = (^) x^12 · 2 xdx = (^2) x dx and v = x 22. Using the techniques of integration by parts, we have ∫ (^2) 1
x ln(x^2 ) dx = ln(x^2 ) · x
2 2
2 1
1
x^2 2 ·^
x dx
= [2 ln 4 − 0] − x
2 2
2 1 = 2 ln 4 − 32.
(10 pts.)(b) Evaluate the indefinite integral ∫ (^2) x (x + 1)(3x − 1) dx.
Write (^) (x+1)(3^2 xx−1) = (^) xA+1 + (^3) xB− 1. Then, 2 x ≡ A(3x − 1) + B(x + 1) ≡ (3A + B)x + (−A + B).
So, 2 = 3A + B and 0 = −A + B. It follows that A = 12 = B. Therefore, ∫ (^2) x (x + 1)(3x − 1) dx^ =
∫ (^) dx x + 1 +
∫ (^) dx 3 x − 1 =^12 ln |x + 1| +^16 ln | 3 x − 1 | + C.
4 EXAM II - MARCH 9, 2007
(10 pts.)(a) Write down the third-degree Taylor polynomial P 3 (x) for f based at x 0 = 1.
The third degree Taylor polynomial is given by
P 3 (x) = f (1) + f ′(1)(x − 1) + f ′′(1) (x^ −^ 1)
2 2! +^ f^
′′′(1) (x^ −^ 1)^3 3! = 1 − 3(x − 1) + (x − 1)^2 + (x^ −^ 1)
3
(5pts.)(b) Suppose it is known that for 0 ≤ x ≤ 2, |f (4)(x)| ≤ 0 .5. What is the maximum possible error committed by using P 3 (x) to estimate f (x) for 0 ≤ x ≤ 2?
The maximum error is given by K 4!^4 |x − 1 |^4 where K 4 is the bound
(5 pts.)(c) Suppose g(x) = f (x + 1). Use part (a) to find the third-degree Maclaurin polynomial for g.
Since g(x) = f (x + 1), it follows that g(k)(x) = f (k)(x + 1). In particular, we have g(k)(0) = f (k)(1). Thus,
M 3 (x) = g(0) + g′(0)x + g
2! x
(^2) + g′′′(0) 3! x
3
= 1 − 3 x + x^2 + x
3
MATH106A CALCULUS II - PROF. P. WONG 5 4.(10 pts.)(a) Let f (x) = cos(2x). Find the fourth-degree Maclaurin polynomial M 4 (x) for f.
First, we have f ′(x) = −2 sin(2x), f ′′(x) = −4 cos(2x), f ′′′(x) = 8 sin(2x) and f (4)(x) = 16 cos(2x). It follows that f (0) = 1, f ′(0) = 0, f ′′(0) = − 4 , f ′′′(0) = 0, and f (4)(0) = 16. Thus,
M 4 (x) = 1 − (^) 2!^4 x^2 +^16 4! x^4 = 1 − 2 x^2 +^23 x^4.
(10 pts.)(b) Solve the following Initial Value Problem y′^ = (1 + x^2 )ey, y(0) = 0.
Separating the variables yields dy ey^ = (1 +^ x
(^2) )dx
or (^) ∫ e−y^ dy =
(1 + x^2 ) dx.
It follows that −e−y^ = x + x 33 + C. The initial condition y(0) = 0 implies that C = − 1 so that e−y^ = 1 − x − x 33. It follows that
y = − ln | 1 − x − x
3 3 |.
