Calculus: Applications of Integration: Area Between Curves Worksheet

Question 1

Find the area enclosed by y = x + 2 and y = x^2.

Accepted Answer

The curves intersect when x^2 = x + 2, so x = -1 and x = 2. On this interval, y = x + 2 is above y = x^2. The area is ∫ from -1 to 2 of (x + 2 - x^2) dx = 9/2 square units.

Question 2

Find the area between y = 4 - x^2 and the x-axis.

Accepted Answer

The curve meets the x-axis at x = -2 and x = 2. The area is ∫ from -2 to 2 of (4 - x^2) dx = 32/3 square units.

Question 3

Find the area enclosed by y = x^2 and y = 2x.

Accepted Answer

The curves intersect at x = 0 and x = 2. On this interval, y = 2x is above y = x^2. The area is ∫ from 0 to 2 of (2x - x^2) dx = 4/3 square units.

Question 4

Set up and evaluate the area between y = e^x and y = 1 on the interval 0 ≤ x ≤ ln 3.

Accepted Answer

On the interval, e^x is above 1. The area is ∫ from 0 to ln 3 of (e^x - 1) dx = 2 - ln 3 square units.

Question 5

Find the area between y = sin x and y = cos x on the interval 0 ≤ x ≤ π/2.

Accepted Answer

The curves intersect at x = π/4, so the integral must be split. The area is ∫ from 0 to π/4 of (cos x - sin x) dx + ∫ from π/4 to π/2 of (sin x - cos x) dx = 2√2 - 2 square units.

Question 6

Find the total area between y = x^3 and y = x on the interval -1 ≤ x ≤ 1.

Accepted Answer

The curves intersect at x = -1, x = 0, and x = 1. The upper function changes at x = 0, so the total area is ∫ from -1 to 0 of (x^3 - x) dx + ∫ from 0 to 1 of (x - x^3) dx = 1/2 square unit.

Question 7

Find the area of the region bounded by x = y^2 and x = 4.

Accepted Answer

The curves meet when y^2 = 4, so y = -2 and y = 2. Using horizontal slices, the area is ∫ from -2 to 2 of (4 - y^2) dy = 32/3 square units.

Question 8

Find the area enclosed by y = 6 - x and y = x^2.

Accepted Answer

The curves intersect when x^2 = 6 - x, so x = -3 and x = 2. The line is above the parabola on this interval. The area is ∫ from -3 to 2 of (6 - x - x^2) dx = 125/6 square units.

Question 9

A student computes ∫ from 0 to 3 of (x^2 - 3x) dx to find the area between y = x^2 and y = 3x. The result is negative. Explain what went wrong and write the correct area integral.

Accepted Answer

The student subtracted in the wrong order. On 0 ≤ x ≤ 3, y = 3x is above y = x^2, so the correct area integral is ∫ from 0 to 3 of (3x - x^2) dx. This gives an area of 9/2 square units.

Question 10

Find the area between y = √x and y = x on the interval 0 ≤ x ≤ 1.

Accepted Answer

On 0 ≤ x ≤ 1, √x is above x. The area is ∫ from 0 to 1 of (√x - x) dx = 1/6 square unit.

Question 11

Find the area between y = x^2 - 4 and the x-axis.

Accepted Answer

The curve intersects the x-axis at x = -2 and x = 2. Between these values, the x-axis is above the curve. The area is ∫ from -2 to 2 of (0 - (x^2 - 4)) dx = ∫ from -2 to 2 of (4 - x^2) dx = 32/3 square units.

Question 12

Set up, but do not evaluate, the integral for the area enclosed by y = x^2 and y = 3.

Accepted Answer

The curves intersect at x = -√3 and x = √3. Since y = 3 is above y = x^2 between those values, the area is ∫ from -√3 to √3 of (3 - x^2) dx.

Question 13

Find the area enclosed by x = y + 2 and x = y^2.

Accepted Answer

The curves intersect when y^2 = y + 2, so y = -1 and y = 2. For these y-values, x = y + 2 is to the right of x = y^2. The area is ∫ from -1 to 2 of ((y + 2) - y^2) dy = 9/2 square units.

Question 14

Suppose f(x) is above g(x) for 1 ≤ x ≤ 4, and f(x) - g(x) = 2x + 1. Find the area between the curves on this interval.

Accepted Answer

Since f(x) is above g(x), the area is ∫ from 1 to 4 of (f(x) - g(x)) dx = ∫ from 1 to 4 of (2x + 1) dx = 18 square units.

Question 15

Find the area enclosed by y = x^2 and y = 4x - x^2.

Accepted Answer

The curves intersect when x^2 = 4x - x^2, so x = 0 and x = 2. On this interval, y = 4x - x^2 is above y = x^2. The area is ∫ from 0 to 2 of ((4x - x^2) - x^2) dx = 8/3 square units.

Calculus: Applications of Integration: Area Between Curves

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