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Calculus is most common subject I know so far. This is one of past exam papers you can find in my uploads. Key points of the exam are: Indefinite, Evaluate, Exact Value, Definite Integral, Compute, Function, Table, Definite Integral, Bounded, Graph
Typology: Exams
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EXAM I - FEBRUARY 4, 2005
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 4, 2005
1.(10 pts.)(a) Evaluate the indefinite integral ∫ (^) sin( 1 t ) t^2 dt. Let u = (^1) t. Then, du = − (^) t^12 dt. By substitution, we have ∫ (^) sin( 1 t ) t^2 dt^ =
− sin u du = cos u + C
= cos
t
(10 pts.)(b) Evaluate the indefinite integral ∫ x cos(2x) dx. Let u = x and dv = cos 2x dx. Then, du = dx and v = 12 sin 2x. Using integration by parts, we have ∫ x cos(2x) dx = x · 12 sin 2x −
2 sin 2x dx = x^ sin 2 2 x− (^12)
sin 2x dx
= x^ sin 2 2 x+ cos 2 4 x+ C.
4 EXAM I - FEBRUARY 4, 2005
3.(15 pts.)(a) The region R is bounded by the graph of y = x^3 , the line x = 1 and the axis y = 0. Find the exact volume of the solid formed by revolving the region R about the x-axis.
∆ x
y = x^3
The exact volume of the solid is given by
V =
0
πy^2 dx
=
0
πx^6 dx = π x
7 7
1 0 = π 7. (5 pts.)(b) Write (DO NOT evaluate) a definite integral representing the arc-length of the path given by y = x^3 from the origin (0,0) to the point (1,1). The curve is given by y = x^3 so that dy dx = 3x^2. The arc length of the desired path is given by
Arc Length =
0
( (^) dy dx
dx
=
0
1 + 9x^4 dx.
MATH106D CALCULUS II - PROF. P. WONG 5 4.(7 pts.)(a) Compute the exact value of the improper integral, if it exists. ∫ (^2) 0
√^ s 4 − s^2
ds ∫ (^2) 0
√^ s 4 − s^2
ds = lim b→ 2
∫ (^) b 0
√^ s 4 − s^2
ds
= lim b→ 2 −
4 − s^2
b 0 = lim b→ 2 (−
4 − b^2 ) − (−
(7 pts.)(b) Compute the exact value of the improper integral, if it exists. ∫ (^) ∞ 0
3 re−r^2 dr ∫ (^) ∞ 0
3 re−r^2 dr = lim b→∞
∫ (^) b 0
3 re−r^2 dr
= lim b→∞ − 32 e−r^2
b 0 = lim b→∞(− 32 e−b^2 ) − (− 32 e^0 ) =^32.
(6 pts.)(c) Determine, by comparison, whether the following improper integral converges or not. Explain briefly. ∫ (^) ∞ 1
e−x^2 dx
For x ≥ 1 , 3 xe−x^2 ≥ e−x^2 ≥ 0. It follows that ∫ (^) ∞ 0
3 xe−x^2 dx ≥
1
3 xe−x^2 dx ≥
1
e−x^2 dx.
By (b),
0 3 xe−x
(^2) dx = 3
1 e−x
(^2) dx converges.
