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Calculus II: Exam Questions from Spring 2004, Math 121, Exams of Calculus

Exam questions from a calculus ii course, math 121, held in spring 2004. The questions cover various topics such as acceleration, area, volume, integration, and differentiation. Students are required to apply concepts related to limits, derivatives, and integrals.

Typology: Exams

Pre 2010

Uploaded on 08/07/2009

koofers-user-olw
koofers-user-olw 🇺🇸

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Math 121 Calculus II Spring 2004
Prob Pts
1
2
3
4
5
6
7
8
9
Total
Final Exam Name: (print neatly)
Instructor: (sign)
1. (5 pts.) Suppose that the acceleration of a particle along the xaxis is given a(t) =
1e2tfor 0 t5, and suppose that at t= 5 we have x(5) = 0 and v(5) = 5.
What is the velocity of the particle at t= 0?
2. (10 pts.) Consider the region sketched below which is bounded by the y-axis, the line
y=x+ 4 and the curve y= 10 x2.
Find the area of the region
-
6
x
1 2
10
1
2
3
4
5
6
7
8
9
1 of 8
pf3
pf4
pf5
pf8

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Prob Pts Math 121^ Calculus II^ Spring 2004

Total

Final ExamInstructor: Name: (print neatly)(sign)

  1. (^1) What is the velocity of the particle at − ( e5 pts.− 2 t (^) for 0) Suppose that the acceleration of a particle along the ≤ t ≤ 5, and suppose that at t = 0? t = 5 we have x(5) = 0 and x axis is given v(5) = 5. a(t) =
  2. ( yFind the area of the region = 10 pts.x + 4 and the curve) Consider the region sketched below which is bounded by the y = 10 − x (^2). y-axis, the line

6

1 2 x

3.generated by rotating the region bounded by the y = 10(10 pts. − x (^2) around the axis) Set up the integral(s) to compute the volume of the solid of revolution x = −1. y-axis, the line y = x + 4 and the curve

6

1 2 x

  1. ( fCompute (2) = 9,4 pts. f ∫) Suppose ′(2) = 2, 2 f f ′′ (0) = 3,(2) = 0. f ′(0) = 0, f ′′(0) = −1, f (1) = 6, f ′(1) = −1, f ′′(1) = −1, 0 f^ ′(x)^ dx^ via the Fundamental Theorem of Calculus.
  1. (the list at the bottom of the page:10 pts. 6 ) Consider each of the graphs below and label it with the correct function from (^51) e - y x

6 (^11) e - y x 6 (^11) e - y x

6 -1 1 1 e -

y x

6 -1^11

e x

y

y y y = 4 + ln(= ln(= ln(xx) (^5) )x) y y y = ln== |e ln(− |xxx|)| y = |ex|

  1. (a. (^) dx15 pts.d [ ln ( ) Evaluate the following derivatives.e 3 x√ (^2) xx + 1 )] =

b. (^) dxd^ (^ e^2 x^ + 2 e−^2 x^ ) =

c) (^) dxd^ [ sin−^1 (3x)]^4 =

c)^ ∫^ √ 1 − 4 x^2 dx

d)^ ∫^ (1^ √−x^ x )dx

e)^ ∫ sec^4 (x) tan^4 (x) dx

f)^ ∫^ x ln(^1 x) dx