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The solutions to exam i for the calculus ii course taught by prof. P. Wong, covering topics such as definite integrals, indefinite integrals, area calculations, and error estimation. Students can use this document as a reference for understanding the concepts and solving similar problems.
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EXAM I - FEBRUARY 6, 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
1
2 EXAM I - FEBRUARY 6, 2009
1.(10 pts.)(a) Find the exact value (by the Fundamental Theorem of Calculus) of the definite integral ∫ (^) e 1
ln(2x) x dx.
Let u = ln(2x). Then du = (^21) x · 2 dx = dxx. When x = 1, u = ln 2; when x = e, u = ln 2e = ln 2 + 1. It follows that ∫ (^) e 1
ln(2x) x dx^ =
∫ (^) ln 2+ ln 2
u du = u
2 2
ln 2+ ln 2 =^12 [(ln 2 + 1)^2 − (ln 2)^2 ] =^12 [2 ln 2 + 1] = ln 2 +^12.
(10 pts.)(b) Evaluate the indefinite integral ∫ xe(4−x^2 )^ dx.
Let w = 4 − x^2. Thus, dw = − 2 x dx or x dx = − 12 dw. Now, ∫ xe(4−x^2 )^ dx = − (^12)
ew^ dw
= − 12 ew^ + C
= − 12 e(4−x^2 )^ + C.
4 EXAM I - FEBRUARY 6, 2009
y
x
1
2
− −
(^1 2 3 4 5 6 7 )
Find R 8 , M 4 using the right-hand sum and the mid-point rule respectively for estimating the definite integral ∫^08 h(x) dx. For R 8 , n = 8 and ∆x = 1 so that R 8 = [2 + 1 + (−2) + (−2) + (−2) + (−1) + (−1) + 1] · ∆x = − 4.
For M 4 , n = 4 and ∆x = 2 so that
M 4 = [2 + (−2) + (−2) + (−1)] · ∆x = − 6.
(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
(cos x)ex^ dx.
Note that f (x) = cos xex. It follows that f ′(x) = cos xex^ − sin xex^ = (cos x − sin x)ex. Thus,
|f ′(x)| ≤ | cos x − sin x|ex^ ≤ | cos x| + | sin x|ex^ ≤ (1 + 1)ex^ = 2ex.
Hence, over the interval [0, 2], |f ′(x)| ≤ 2 e^2 so that we can choose K 1 = 2e^2. The error committed by L 10 must be less than or equal to 2 e
(^2) (2−0) 2 20 =^25 e^2.
MATH106A,B CALCULUS II - PROF. P. WONG 5
B
A
y=
x=y^2
x = −
Using horizontal slices each of which looks like a washer, the volume of the solid of revolution is given by
V =
0
π(y^2 + 1)^2 − π(1)^2 dy
= π
0
y^4 + 2y^2 dy
= π
( (^) y 5 5 +
2 y^3 3
176 π
(5 pts.) SET UP (do not evaluate) a definite integral representing the arc length of the portion of the curve x = y^2 from A to B [Hint: What is f (x) here?]. The point A has coordinates (0, 0) and B has coordinates (4, 2). Here, the portion of the curve in question is the graph of f (x) = y = √x from x = 0 to x = 4. Since f ′(x) = 2 √^1 x , the length of this path is given by
L =
0
1 + [f ′(x)]^2 dx
=
0
1 + (^41) x dx.
