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Solutions to exam ii for calculus ii (math106a,b) taught by professor p. Wong. It includes the evaluation of indefinite integrals using various techniques such as integration by parts, substitution, and completing the square. Additionally, it covers finding the third-degree taylor polynomials for given functions.
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EXAM II - MARCH 7, 2008
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 7, 2008
1.(10 pts.)(a) Evaluate the indefinite integral ∫ x^2 ln(2x) dx.
Let u = ln(2x) and dv = x^2 dx. Thus, du = (^21) x · 2 dx = dxx and v = x 33.
Using the technique of integration by parts, ∫ x^2 ln(2x) dx = (ln(2x)) ·
x^3 3
x^3 3
· dx x
= x
3 3
ln(2x) − 1 3
x^2 dx
= x
3 3
ln(2x) − x
3 9
(10 pts.)(b) Evaluate the indefinite integral ∫ (^) x (^3) + 3 (x + 1)x
dx.
Applying long division, we obtain x^3 + 3 (x + 1)x = (x^ −^ 1) +^
x + 3 (x + 1)x.
Now, we use partial fractions for (^) (xx+1)+3x. First, we write
x + 3 (x + 1)x =^
(x + 1) +^
x.
Therefore, Ax + B(x + 1) ≡ x + 3. It follows that A + B = 1 and B = 3
and so A = − 2. It follows that ∫ x^3 + 3 (x + 1)x
dx =
(x − 1) dx +
x + 3 (x + 1)x
dx
=
(x − 1) dx +
(x + 1)
dx +
x
dx
= x
2 2
− x − 2 ln |x + 1| + 3 ln |x| + C.
4 EXAM II - MARCH 7, 2008
rin polynomial M 3 (x) for f.
Since f (x) = arctan x, we have f ′(x) = (^) 1+^1 x 2 = (1 + x^2 )−^1 , f ′′(x) = −(1 + x^2 )−^2 · 2 x and f ′′′(x) = [2(1 + x^2 )−^3 · 2 x] · 2 x + [−(1 + x^2 )−^2 ] · 2. Now,
f (0) = 0, f ′(0) = 1, f ′′(0) = 0 and f ′′′(0) = − 2.
It follows that
M 3 (x) = f (0) + f ′(0)x +
f ′′(0) 2! x
(^2) + f^ ′′′(0) 3! x
3
= x −
3! x
(^3) = x − x^3
(10 pts.)(b) Let g(x) =
x. Find the third-degree Taylor polynomial
P 3 (x) for g(x) centered at x 0 = 4.
Since g(x) =
x = x^1 /^2 , we have g′(x) = 12 x−^1 /^2 , g′′(x) = − 14 x−^3 /^2 and g′′′(x) = 38 x−^5 /^2. Now
g(4) = 2, g′(4) =^1 4
, g′′(4) = − 1 32
and g′′′(x) =^3 8
It follows that
P 3 (x) = g(4) + g′(4)(x − 4) +
g′′(4) 2! (x^ −^ 4)
(^2) + g′′′(4) 3! (x^ −^ 4)
3
= 2 +^14 (x − 4) − 64 1 (x − 4)^2 + 512 1 (x − 4)^3.
MATH106A,B CALCULUS II - PROF. P. WONG 5
4.(10 pts.)(a) Let f (x) = cos(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 − 2 ≤ x ≤ 2?
Since f (x) = cos(2x), then f ′(x) = −2 sin(2x), f ′′(x) = −4 cos(2x), f ′′′(x) =
8 sin(2x), and f (4)(x) = 16 cos(2x). Note that |f (4)(x)| ≤ 16 so we can
choose K 4 = 16. Over the interval [− 2 , 2], |x| ≤ 2. Therefore, Tay-
lor’s theorem asserts that the maximum error committed by M 3 (x)
is given by K 4 |x|^4 4!
4 4!
(10 pts.)(b) Let
h(x) =
k sin x, if 0 ≤ x ≤ π; 0 , otherwise.
Find k for which h(x) is a probability density function.
For h(x) to be a probability density function, h(x) ≥ 0 for all x.
Since sin x ≥ 0 for 0 ≤ x ≤ π, this implies that k ≥ 0. In addition, ∫ (^) ∞ −∞ h(x)^ dx^ must be equal to^1. It follows that
1 =
−∞
h(x) dx =
∫ (^) π
0
k sin x dx = −k cos x
∣∣π 0 =^ −k(−1)^ −^ (−k).
This means that 2 k = 1 or k = 12.
