Practice Set B: Applications of Integration | AP Calculus AB

Question 1

The area of a region bounded on the left by x = g(y) and on the right by x = f(y) from y = c to y = d is:

Solution: Using horizontal strips, area = integral of (right curve minus left curve) with respect to y: integral from c to d of [f(y) - g(y)] dy.

Question 2

If a solid's cross-sectional area is A(x) = x^2 on [0, 2], its volume is:

Solution: V = integral from 0 to 2 of x^2 dx = [x^3/3] from 0 to 2 = 8/3.

Question 3

Find the area under y = 1/x from x = 1 to x = e.

Solution: Area = integral from 1 to e of (1/x) dx = [ln x] from 1 to e = ln(e) - ln(1) = 1.

Question 4

A particle has v(t) = 4t - t^2. On [0, 4], the velocity is nonnegative, so the total distance equals:

Solution: Since v >= 0 on [0, 4], distance = integral from 0 to 4 of (4t - t^2) dt = [2t^2 - t^3/3] from 0 to 4 = 32 - 64/3 = 32/3.

Question 5

The volume of a solid with square cross sections is found by integrating:

Solution: Volume by cross sections integrates the cross-sectional area; for squares that area is side^2.

Question 6

Find the area bounded by x = y^2 and x = 4 (with respect to y).

Solution: Meets 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 7

Rotating the region under y = x + 1 on [0, 2] about the x-axis, the disk radius at position x is:

Solution: For rotation about the x-axis, the radius equals the function value, x + 1.

Question 8

The average value of f(x) = 4 - x on [0, 4] is:

Solution: Average = (1/4) integral from 0 to 4 of (4 - x) dx = (1/4)[4x - x^2/2] from 0 to 4 = (1/4)(16 - 8) = 2.

Question 9

Find the area between the curves y = sin(x) and y = cos(x) from x = 0 to their first intersection at x = pi/4.

Solution: On [0, pi/4], cos(x) >= sin(x). Area = integral from 0 to pi/4 of (cos x - sin x) dx = [sin x + cos x] from 0 to pi/4 = (sqrt(2)) - (1) = sqrt(2) - 1. (Adapted from the Vidaara NCERT/JEE bank.)

Question 10

Find the area of the region bounded by y = e^x, y = e^(-x), and the vertical line x = 1.

Solution: For x in [0, 1], e^x >= e^(-x). Area = integral from 0 to 1 of (e^x - e^(-x)) dx = [e^x + e^(-x)] from 0 to 1 = (e + e^(-1)) - (1 + 1) = e + e^(-1) - 2. (Adapted from the Vidaara NCERT/JEE bank.)

Question 11

Find the area of the region bounded by the parabola y^2 = x and the line y = x - 2 (integrating with respect to y).

Solution: The curves meet at y = -1 and y = 2 (x = y^2 and x = y + 2). Area = integral from -1 to 2 of [(y + 2) - y^2] dy = [y^2/2 + 2y - y^3/3] from -1 to 2 = 10/3. (Adapted from the Vidaara NCERT/JEE bank.)

Question 12

Find the area bounded by y = ln(x), y = 0, and x = e (using the region under the curve).

Solution: Area = integral from 1 to e of ln(x) dx = [x ln x - x] from 1 to e = (e - e) - (0 - 1) = 1. (Adapted from the Vidaara NCERT/JEE bank.)

Question 13

A solid has a base bounded by y = x and y = x^2 on [0, 1], with square cross sections perpendicular to the x-axis. Which integral gives the volume?

Solution: The side of each square equals the vertical distance between the curves, (x - x^2). The area is (x - x^2)^2, so V = integral from 0 to 1 of (x - x^2)^2 dx.

Question 14

A solid has semicircular cross sections (diameter = height of the region) perpendicular to the x-axis, with region height h(x). The cross-sectional area is:

Solution: A semicircle with diameter h(x) has radius h(x)/2. Area = (1/2)pi r^2 = (1/2)pi(h/2)^2 = (pi/8) h^2.

Question 15

Find the volume of the solid generated by rotating the region under y = x^3 on [0, 1] about the x-axis.

Solution: V = pi integral from 0 to 1 of (x^3)^2 dx = pi integral from 0 to 1 of x^6 dx = pi[x^7/7] from 0 to 1 = pi/7. (Adapted from the Vidaara NCERT/JEE bank.)

Question 16

Find the volume when the region under y = sqrt(a^2 - x^2) from x = -a to x = a is rotated about the x-axis.

Solution: V = pi integral from -a to a of (a^2 - x^2) dx = pi[a^2 x - x^3/3] from -a to a = (4/3)pi a^3 (the volume of a sphere of radius a). (Adapted from the Vidaara NCERT/JEE bank.)

Question 17

Find the volume when the region bounded by y = 2x - x^2 and the x-axis is rotated about the x-axis.

Solution: The region spans x = 0 to x = 2. V = pi integral from 0 to 2 of (2x - x^2)^2 dx = pi integral of (4x^2 - 4x^3 + x^4) dx = pi[4x^3/3 - x^4 + x^5/5] from 0 to 2 = 16pi/15. (Adapted from the Vidaara NCERT/JEE bank.)

Question 18

Rotate the region between y = x and y = x^2 on [0, 1] about the x-axis. The volume is:

Solution: Outer radius x, inner x^2. V = pi integral from 0 to 1 of (x^2 - x^4) dx = pi[x^3/3 - x^5/5] from 0 to 1 = pi(1/3 - 1/5) = pi(2/15) = 2pi/15.

Question 19

A solid has a base that is the region under y = 4 - x^2 (x from -2 to 2, y >= 0) and equilateral-triangle cross sections perpendicular to the x-axis with side (4 - x^2). The area of each cross section is:

Solution: An equilateral triangle of side s has area (sqrt(3)/4)s^2. With s = 4 - x^2, the area is (sqrt(3)/4)(4 - x^2)^2.

Question 20

Using the Shell Method, find the volume when the region under y = x on [0, 2] is rotated about the y-axis.

Solution: V = 2 pi integral from 0 to 2 of x(x) dx = 2 pi integral from 0 to 2 of x^2 dx = 2 pi[x^3/3] from 0 to 2 = 2 pi(8/3) = 16pi/3.