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The solutions to exam i for calculus ii, taught by prof. P. Wong, dated february 1, 2008. Problems related to definite and indefinite integrals, finding areas, and estimating definite integrals using the right-hand sum and mid-point rule. It also covers topics such as finding the volume of a solid obtained from rotating a region around an axis, and estimating the value of a solution using euler's method.
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EXAM I - FEBRUARY 1, 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
1
2 EXAM I - FEBRUARY 1, 2008
1.(10 pts.)(a) Find the exact value (by the Fundamental Theorem of Calculus) of the definite integral ∫ (^) e 1
1 + (ln x)^2 x dx.
Let u = ln x. Then du = (^) x^1 dx. When x = 1, u = 0 and when x = e, u = 1. Thus, (^) ∫ (^) e
1
1 + (ln x)^2 x dx^ =
0
1 + u^2 du
= u + u
3 3
1 0 =
(10 pts.)(b) Evaluate the indefinite integral ∫ (^) x √ 1 − x^4
dx.
Let u = x^2. Then du = 2x dx or x dx = 12 dx. Thus, ∫ (^) x √ 1 − x^4
dx =^12
∫ (^) du √ 1 − u^2 =^12 arcsin u + C =^12 arcsin(x^2 ) + C.
4 EXAM I - FEBRUARY 1, 2008
1 h(x)^ dx.
R 6 = [h(1.5) + h(2) + h(2.5) + h(3) + h(3.5) + h(4)] · ∆x = [2 + 1 + 0 + (−2) + 3 + 1] · (0.5) = 2. 5 and
M 3 = [h(1.5) + h(2.5) + h(3.5)] · ∆x (here the ∆x is 1) = [2 + 0 + 3] · (1) = 5
(10 pts.)(b) Recall that the error committed by using the left hand sum approximation Ln is less than or equal to K^1 ·( 2 bn− a)^2 where |f ′(x)| ≤ K 1 for some constant K 1 over the interval [a, b]. Use this result to give an upper bound for the error committed by L 10 for
I =
0
(sin x)ex^ dx.
Here, f (x) = sin x ex^ so f ′(x) = cos x ex^ + sin x ex^ = ex(sin x + cos x). Now, |f ′(x)| = ex| sin x + cos x| ≤ ex(1 + 1). Over the interval [0, 2], |f ′(x)| ≤ 2 e^2 so we can choose K 1 = 2e^2. It follows that
|I − L 10 | ≤ 2 e
2 e^2
One can also use the calculator to estimate that |f ′(x)| < 5 (note that the maximum of f ′(x) does NOT occur at x = 2) so that one can choose K 1 to be 5.
MATH106A,B CALCULUS II - PROF. P. WONG 5
A
B y=−1−x
x=
2
A typical slice of this solid is a circular disk with thickness ∆x so that the integral for the volume is with respect to x. The volume is given by
V =
0
π(− 1 − x^2 )^2 dx = π
0
1 + 2x^2 + x^4 dx = π
x +^2 x
3 3 +^
x^5 5
0 = π
=^2815 π.
(5 pts.) SET UP (do not evaluate) a definite integral representing the arc length of the portion of the curve from A to B. The point A has coordinates (0, −1) and B has coordinates (1, −2). The arc length of the path from A to B is given by
L =
0
1 + [f ′(x)]^2 dx =
0
1 + [− 2 x]^2 dx (where f (x) = − 1 − x^2 )
=
0
1 + 4x^2 dx.
