Multiple Choice
Question 1
The area bounded by x = y^2 and x = 4 (integrating with respect to y) is:
A
32/3
B
16/3
C
64/3
D
8
Solution: The curves meet at y = -2 and y = 2. Area = integral from -2 to 2 of (4 - y^2) dy = [4y - y^3/3] from -2 to 2 = 32/3.
Question 2
A solid has a base region and square cross sections perpendicular to the x-axis with side length s(x). Its volume is:
A
integral of [s(x)]^2 dx
B
integral of s(x) dx
C
pi integral of [s(x)]^2 dx
D
integral of 4 s(x) dx
Solution: For square cross sections, the cross-sectional area is s(x)^2, so the volume is the integral of s(x)^2 over the base interval.
Question 3
By the Disk Method, rotating the region under y = f(x) on [a, b] about the x-axis gives volume:
A
pi integral from a to b of [f(x)]^2 dx
B
integral from a to b of [f(x)]^2 dx
C
2 pi integral from a to b of f(x) dx
D
pi integral from a to b of f(x) dx
Solution: Each disk has radius f(x) and area pi[f(x)]^2, so V = pi integral from a to b of [f(x)]^2 dx.
Question 4
By the Washer Method, rotating the region between y = R(x) (outer) and y = r(x) (inner) about the x-axis gives:
A
pi integral of ([R(x)]^2 - [r(x)]^2) dx
B
pi integral of (R(x) - r(x))^2 dx
C
pi integral of (R(x) - r(x)) dx
D
2 pi integral of R(x) dx
Solution: A washer's area is pi(R^2 - r^2), so the volume is pi integral of ([R(x)]^2 - [r(x)]^2) dx.
Question 5
For a solid where the cross-sectional area is given by A(x), the volume over [a, b] is:
A
integral from a to b of A(x) dx
B
pi integral from a to b of A(x) dx
C
integral from a to b of [A(x)]^2 dx
D
A(b) - A(a)
Solution: Volume by known cross sections is the integral of the cross-sectional area function: integral from a to b of A(x) dx.
Question 6
A solid's cross-sectional area is A(x) = 4x. What is its volume from x = 0 to x = 3?
A
18
B
12
C
36
D
9
Solution: V = integral from 0 to 3 of 4x dx = [2x^2] from 0 to 3 = 18.
Question 7
By the Shell Method, rotating the region under y = f(x) on [a, b] about the y-axis gives volume:
A
2 pi integral from a to b of x f(x) dx
B
pi integral from a to b of [f(x)]^2 dx
C
2 pi integral from a to b of f(x) dx
D
pi integral from a to b of x f(x) dx
Solution: A cylindrical shell has radius x, height f(x), and thickness dx, so its volume element is 2 pi x f(x) dx; V = 2 pi integral from a to b of x f(x) dx.
Question 8
Find the average value of f(x) = 3x^2 on [0, 2].
A
4
B
8
C
12
D
2
Solution: Average = (1/2) integral from 0 to 2 of 3x^2 dx = (1/2)[x^3] from 0 to 2 = (1/2)(8) = 4.
Question 9
Find the area between y = x^2 and y = x from x = 0 to x = 1.
A
1/6
B
1/2
C
1/3
D
5/6
Solution: On [0, 1], x >= x^2. Area = integral from 0 to 1 of (x - x^2) dx = 1/2 - 1/3 = 1/6.
Question 10
Rotate the region under y = 2 (a constant) on [0, 3] about the x-axis. The volume (a cylinder) is:
A
12pi
B
6pi
C
4pi
D
9pi
Solution: V = pi integral from 0 to 3 of (2)^2 dx = pi(4)(3) = 12pi (a cylinder of radius 2 and height 3).
Question 11
A car's velocity is v(t) = 60 (constant, mph) for 2 hours. The distance traveled is:
A
120 miles
B
60 miles
C
30 miles
D
62 miles
Solution: Distance = integral from 0 to 2 of 60 dt = 60(2) = 120 miles.
Question 12
The region bounded by x = y^2 and x = y + 2 is most naturally integrated:
A
with respect to y
B
with respect to x only
C
using polar coordinates
D
it cannot be found
Solution: Since both boundaries are given as x in terms of y, integrating with respect to y (horizontal strips) is most direct.
Question 13
Rotating the region under y = x from x = 0 to x = 3 about the x-axis produces a cone with volume:
A
9pi
B
27pi
C
3pi
D
6pi
Solution: V = pi integral from 0 to 3 of x^2 dx = pi[x^3/3] from 0 to 3 = pi(9) = 9pi.
Question 14
The average value of a constant function f(x) = 7 on any interval is:
A
7
B
0
C
depends on the interval
D
7 times the length
Solution: The average value of a constant is the constant itself: 7.
Question 15
A particle's acceleration is a(t) = 2, with v(0) = 1 and s(0) = 0. Its position at t = 2 is:
A
6
B
4
C
5
D
2
Solution: v(t) = 2t + 1, s(t) = t^2 + t. At t = 2, s = 4 + 2 = 6.
Question 16
Find the area between y = 4 - x^2 and the x-axis.
A
32/3
B
16/3
C
8
D
64/3
Solution: The parabola meets the x-axis at x = -2 and x = 2. Area = integral from -2 to 2 of (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = 32/3.
Question 17
Rotating the region under y = f(x) about the x-axis, the radius of each disk equals:
A
f(x)
B
x
C
[f(x)]^2
D
2 f(x)
Solution: For rotation about the x-axis, the disk's radius is the distance from the axis to the curve, which is f(x).
Question 18
The Shell Method is most convenient when rotating a region about the y-axis and the region is described by:
A
y as a function of x
B
x as a function of y only
C
a circle
D
a constant
Solution: Shells (radius x, height f(x)) work naturally when the curve is y = f(x) and rotation is about the y-axis, avoiding solving for x.
Question 19
For a solid with circular (disk) cross sections of radius r(x), the cross-sectional area is:
A
pi [r(x)]^2
B
2 pi r(x)
C
[r(x)]^2
D
pi r(x)
Solution: A circle of radius r(x) has area pi[r(x)]^2.
Question 20
The average value of f(x) = x on [0, 4] is:
A
2
B
4
C
8
D
1
Solution: Average = (1/4) integral from 0 to 4 of x dx = (1/4)[x^2/2] from 0 to 4 = (1/4)(8) = 2.