Calculus: Applications of Integration: Area Between Curves
Set up and evaluate definite integrals to find enclosed area
Calculus: Applications of Integration: Area Between Curves
Set up and evaluate definite integrals to find enclosed area
Math - Grade 9-12
- 1
Find the area enclosed by y = x + 2 and y = x^2.
First solve x^2 = x + 2 to find the limits of integration.
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. - 2
Find the area between y = 4 - x^2 and the x-axis.
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. - 3
Find the area enclosed by y = x^2 and y = 2x.
Use a test value such as x = 1 to decide which graph is on top.
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. - 4
Set up and evaluate the area between y = e^x and y = 1 on the interval 0 ≤ x ≤ ln 3.
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. - 5
Find the area between y = sin x and y = cos x on the interval 0 ≤ x ≤ π/2.
The upper function changes at x = π/4.
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. - 6
Find the total area between y = x^3 and y = x on the interval -1 ≤ x ≤ 1.
Because the curves cross inside the interval, use absolute area or split the integral.
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. - 7
Find the area of the region bounded by x = y^2 and x = 4.
Integrate with respect to y because the curves are written as x-values.
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. - 8
Find the area enclosed by y = 6 - x and y = x^2.
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. - 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.
Area is found with upper function minus lower function.
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. - 10
Find the area between y = √x and y = x on the interval 0 ≤ x ≤ 1.
On 0 ≤ x ≤ 1, √x is above x. The area is ∫ from 0 to 1 of (√x - x) dx = 1/6 square unit. - 11
Find the area between y = x^2 - 4 and the x-axis.
If the curve is below the x-axis, subtract the curve from 0.
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. - 12
Set up, but do not evaluate, the integral for the area enclosed by y = x^2 and y = 3.
Find the x-values where x^2 = 3.
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. - 13
Find the area enclosed by x = y + 2 and x = y^2.
Use right minus left when integrating with respect to y.
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. - 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.
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. - 15
Find the area enclosed by y = x^2 and y = 4x - x^2.
Combine like terms in the integrand before integrating.
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.