Practice Integration Problems #2 Name:______________________________
y
1) The figure to the right shows the graph of the velocity of a particle
moving along the x-axis as a function of time. If the particle is at the A B
origin when t = 0, then which of the marked points is the particle C t
furthest from the origin? E
(A) A (B) B (C) C
(D) D (E) E D
2) sin(3x 4)dx
1
(A) cos(3x 4) C (B) cos(3x 4) C (C) 3cos(3x 4) C
3
1
(D) cos(3x 4) C (E) cos(3x 4) C
3
x, x 0 1
3) Let f(x) be the function defined by f(x) =
x 1, x > 0
. The value of xf ( x)dx
2
(A) 3/2
(B) 5/2
(C) 3
(D) 7/2
(E) 11/2
1
4) The average value of the function f(x) = cos x on the closed interval [-4, 0] is
2
(A) -1/2sin(2) (B) -1/4sin(2) (C) 1/2cos(2) (D) 1/4sin(2) (E) 1/2sin(2)
5) Let R(t) represent the rate at which water is leaking out of a tank, where t is measured in hours. Which of the
following expressions represents the total amount of water in gallons that leaks out in the first three hours?
R(3) R(0)
3 3 3
1
(A) R(3) – R(0) (B)
30
(C) R(t )dt
0
(D) R '(t )dt
0
(E)
3 0
R(t )dt
1 7 1
6) Suppose that f(x) is an even function and let
0
f ( x)dx = 5 and
0
f ( x)dx = 1. What is f ( x)dx ?
7
(A) – 5
(B) -4
(C) 0
(D) 4
(E) 5
7) As shown in the figure to the right, the function f(x) consists of
a line segment from (0, 4) to (8, 4) and one-quarter of a circle with
a radius of 4. What is the average (mean) value of this function
on the interval [0, 12]? (calc.)
(A) 2
(B) 3.714
(C) 3.9
(D) 22.283
(E) 41.144
8) If f is the function defined by f(x) = 3
x 2 4 x and g is an antiderivative of f such that g(5) = 7, then g(1) ≈ (calc.)
(A) –3.882
(B) –3.557
(C) 1.710
(D) 3.557
(E) 3.882
9) If f and g are continuously differentiable functions defined for all real numbers, which of the following definite
integrals is equal to f(g(4)) – f(g(2))?
4 4 4
(A)
2
f '( g ( x))dx (B)
2
f ( g ( x)) f '( x)dx (C) f ( g ( x)) g '( x)dx
2
4 4
(D) f '( g ( x)) g '( x)dx
2
(E) f '( g '( x)) g '( x)dx
2
x 1
5
10) If the substitution u = x 1 is made, the integral 2
x
dx =
5 2 2
2u 2 u2 u2
(A) 2 u 2 1 du (B) 1 u 2 1 du (C) 1 2(u 2 1) du
5 2
u 2u 2
(D) 2 du (E) 1 u 2 1 du
2
u 1
(2 x kx 2 2k )dx 12 , then k must be
3
11) If
0
(A) -3
(B) -2
(C) 1
(D) 2
(E) 3
x3
d
12)
dx x
sin(t 2 ) dt
(A) sin(x6) – sin(x2) (B) 6x2sin(x3) – 2sin(x) (C) 3x2sin(x6) – sin(x2)
(D) 6x5sin(x6) – 2sin(x2) (E) 2x3cos(x6) –2 cos(x2)
2013 Free Response Question 1 (calc.)