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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: Determining, Indefinite Integral, Exact Value, Definite Integral, Picture, Region Bounded, Points, Intersection, Curves, Region Described
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EXAM I - SEPTEMBER 30, 2011
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
2 EXAM I - SEPTEMBER 30, 2011
1.(10 pts.)(a) Evaluate the indefinite integral (be sure to show all your work) ∫ ex^ sin(ex) dx.
Let u = ex^ so that du = exdx∫. It follows that
ex^ sin(ex) dx =
sin u du = − cos u + C = − cos(ex) + C.
(10 pts.) (b) Find the exact value of the definite integral (be sure to show all your work) ∫ (^2)
0
x^2 √ x^3 + 1
dx.
Let u = x^3 + 1 so that du = 3x^2 dx or x^2 dx = 13 du. When x = 0, u = 1 and when x = 2, u = 9. It follows that ∫ (^2)
0
x^2 √ x^3 + 1
dx =
1
1 / 3 du √ u
1
u−^1 /^2 du
u^1 /^2 (1/2)
9
1
u
9
4 EXAM I - SEPTEMBER 30, 2011
x -2 -1 0 1 2 3 4 g(x) 1 0 -1 -2 1 3 4
Find L 6 , M 3 using the left-hand sum and the mid-point rule respectively for estimating the definite integral
− 2 g(x)^ dx. For L 6 , ∆x = 1 so that
L 6 = [(1) + (0) + (−1) + (−2) + (1) + (3)] · (1) = 2.
For M 3 , ∆x = 2 so that
M 3 = [(0) + (−2) + (3)] · (2) = 2.
First, the derivative f ′(x) = (−e−x) sin x + e−x^ cos x = e−x(cos x − sin x). Thus the required arc length is given by the following definite integral ∫ (^) π/ 2 √ 1 + e−^2 x(cos x − sin x)^2 dx.
MATH106B,C CALCULUS II - PROF. P. WONG 5
(10 pts.) (a) Set up (do not evaluate) a definite integral representing the volume of the solid obtained from rotating the region R around the line x = 0, i.e., the y-axis. [Hint: sketch a picture of the region R first.]
y=x
2(x-2)=y 2
y=
The volume of the solid in question is given by
∫ (^2)
0
π(2 + y^2 2 )^2 − πy^2 dy.
(10 pts.) (b) Find the exact volume of the solid described in part (a).
∫ (^2)
0
π
y^2 2
− πy^2 dy = π
0
4 + 2y^2 + y^4 4
− y^2 dy
= π
0
4 + y^2 +
y^4 4 dy
= π
4 y + y^3 3
y^5 20
0 = π
184 π .
