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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: Definite Integral, Evaluate, Indefinite Integral, Area, Region Bounded, Curve, Line, Determining, Points, Intersections
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EXAM II - MARCH 13, 2009
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 - MARCH 13, 2009
1.(10 pts.)(a) Evaluate the indefinite integral ∫ x^3 4 + x^2
dx.
Long division yields x^3 4 + x^2
= x − 4 x 4 + x^2
Thus, ∫ (^) x 3 4 + x^2 dx^ =
x dx −
∫ (^4) x 4 + x^2 dx
=
x^2 2 −^2
u du^ where^ u^ = 4 +^ x
(^2) and du = 2x dx
x^2 2 −^ 2 ln(4 +^ x
Remark: one can do this problem by a simple substitution u = 4+x^2 without
using long division.
(10 pts.)(b) Evaluate the indefinite integral ∫ x^2 + 3x + 1 x(x^2 + 1) dx. Here, we use the technique of partial fractions. First, write x^2 + 3x + 1 x(x^2 + 1)
x
It follows that A(x^2 + 1) + x(Bx + C) ≡ x^2 + 3x + 1. By equating the
coefficients of the like terms, we obtain A = 1, B = 0, C = 3. Thus, ∫ (^) x (^2) + 3x + 1 x(x^2 + 1) dx^ =
x +^
x^2 + 1 dx = ln |x| + 3 arctan x + C.
4 EXAM II - MARCH 13, 2009
polynomial M 3 (x) for f.
Note that f ′(x) = e−x^ + xe−x(−1) = e−x(1 − x), f ′′(x) = −e−x(1 − x) +
e−x(−1) = e−x(x − 2), and f ′′′(x) = −e−x(x − 2) + e−x(1) = e−x(3 − x). Thus, f (0) = 0, f ′(0) = 1, f ′′(0) = − 2 , and f ′′′(0) = 3. It follows that
M 3 (x) =
k=
f (k)(0) k! x
k (^) = x − x (^2) + x^3
(10 pts.)(b) Let g(x) = ln x. Find the fourth-degree Taylor polynomial
P 4 (x) for g(x) centered at x 0 = 1.
Note that g′(x) = (^1) x = x−^1 , g′′(x) = −x−^2 , g′′′(x) = 2x−^3 , and
g(4)(x) = − 6 x−^4. Thus, g(1) = 0, g′(1) = 1, g′′(1) = − 1 , g′′′(1) = 2, g(4)(1) =
− 6. It follows that
P 4 (x) =
k=
g(k)(1) k!
xk^ = (x − 1) − 1 2
(x − 1)^2 +^1 3
(x − 1)^3 − 1 4
(x − 1)^4.
MATH106A,B CALCULUS II - PROF. P. WONG 5
4.(10 pts.)(a) Let f (x) = sin(2x). What is the maximum possible er-
ror, according to Taylor’s theorem, committed by using the third-degree
Maclaurin polynomial M 3 (x) to estimate f (x) for − 12 ≤ x ≤ 12?
Since f (x) = sin(2x), we have f ′(x) = 2 cos(2x), f ′′(x) = −4 sin(2x), f ′′′(x) = −8 cos(2x), and f (4)(x) = 16 sin(2x). It follows that |f (4)(x)| ≤
16 so we can choose K 4 = 16. Over the interval − 12 ≤ x ≤ 12 , |x|^4 ≤ 161.
Therefore, by Taylor’s theorem,
|f (x) − M 3 (x)| ≤
K 4 |x|^4 4! ≤^
(10 pts.)(b) Let
f (x) =
see graph below, if 0 ≤ x ≤ 4; 0 , otherwise. Find a for which f (x) is a probability density function. Justify your answer.
x
y
(^014)
a
For f (x) to be a probability density function, (i) f (x) ≥ 0 for all
x and (ii)
−∞ f^ (x)^ dx^ = 1.^ Note that (i) is satisfied.^ For (ii), it suffices to show that the area under the graph of f is equal to 1
or the area of the triangle above is 1. This means that 42 a = 1 or
a = 12.
