Midterm 4 Sample C ... Solution: esin(4x), because of the Fundamental Theorem
of Calculus I. b. ... Solution: See Example 2 in Section 4.7 of your textbook.
Midterm 4 Sample C
Loyola University Chicago Math 161-005, Fall 2012
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Please do not start working until instructed to do so. You have 75 minutes. You must show your work to receive full credit. No calculators. You may use one one-sided 8.5 by 11 sheet of handwritten (by you) notes.
Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Problem 6. Total.
Problem 1.(20 points) Find the following integrals. Put a box around your final answer . Z a.(5 points) 3x4 + 7 cos(2x) dx, Solution:
b.(5 points)
Z
1
Z
2e3y dy e3y + 16
Z
π 5
√
0
1 dx 1 − x2
Solution:
c.(5 points) Solution:
d.(5 points)
0
Solution:
sin(5t) dt 9 − cos(5t)
Problem 2.(10 points) Write the right-endpoint Riemann sum for f (x) = x3 − 6xR2 on [1, 3] with 3 n = 4 subintervals. Is this Riemann sum an underestimate or an overestimate of 1 x3 − 6x2 dx? Explain. Solution:
The function f (x) = x3 − 6x2 is decreasing because f 0 (x) = 3x2 − 12x = 3x(x − 4) < 0 for x in [1, 3]. Hence, the right-endpoint Riemann sum is an underestimate.
Problem 3.(10 points) Find the derivatives: Z x d a.(5 points) esin(4t) dt, dx 2 Solution: esin(4x), because of the Fundamental Theorem of Calculus I.
b.(5 points)
d dx
Z
a
arctan(4x) dx, where a is a constant.
6x
Solution: Z 6x Z 6x Z a d d d arctan(4x) dx = − arctan(4x) dx = − arctan(4x) dx = − arctan(24x)6 dx 6x dx dx a a because of the Fundamental Theorem of Calculus I and Chain Rule.
Problem 4.(10 points) Use Newton’s Method (at least 2 steps) to estimate
√
2.
Solution: See Example 2 in Section 4.7 of your textbook.
Problem 5.(10 points) The velocity of a particle along the x-axis, for times t ≥ 0, is given by 3
v(t) = t2 e−t . The position of the particle at time t = 0 is x(t) = 2. Find the position of the particle at time t = 2. Solution: The displacement between t = 0 and t = 2 is Z
0
2
2 −t3
t e
1 dt = − 3
Z
−23
1 1 1 0 e du = − eu |−8 e − e−8 = 0 = 3 3 3 u
0
Hence, the position at time t = 2 is 1 x(2) = x(0) + 3
