Area Between Two Curves Practice Worksheet

Find the area bounded by y = 4 - x^2 and y = 0.

Intersections at x = -2 and x = 2. Integral from -2 to 2 of (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 32/3 square units.

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

Intersections: x^2 = 2x gives x = 0 and x = 2. On [0, 2], 2x is above x^2. Integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2 = (4 - 8/3) - 0 = 4/3 square units.

Find the area between y = x^2 + 1 and y = x + 3.

Intersections: x^2 + 1 = x + 3 leads to x^2 - x - 2 = 0, so x = -1 and x = 2. Line is upper function. Integral from -1 to 2 of ((x + 3) - (x^2 + 1)) dx = Integral from -1 to 2 of (-x^2 + x + 2) dx = [-x^3/3 + x^2/2 + 2x] from -1 to 2 = 9/2 square units.

Find the area of the region bounded by y = x^2 and y = sqrt(x).

Intersections: x^2 = sqrt(x) gives x = 0 and x = 1. On [0, 1], sqrt(x) is above x^2. Integral from 0 to 1 of (x^(1/2) - x^2) dx = [2/3 x^(3/2) - x^3/3] from 0 to 1 = 2/3 - 1/3 = 1/3 square units.

Find the total area enclosed between y = x^3 and y = x.

Intersections at x = -1, 0, 1. From -1 to 0, x^3 is above x: integral of (x^3 - x) dx = 1/4. From 0 to 1, x is above x^3: integral of (x - x^3) dx = 1/4. Total area = 1/4 + 1/4 = 1/2 square units.

Find the area bounded by y = x^2, the tangent line to the curve at x = 2, and the x-axis.

Tangent at (2, 4): derivative is 2x, so slope is 4; equation is y - 4 = 4(x - 2), or y = 4x - 4. Tangent hits x-axis at x = 1. Area split at x = 1: integral from 0 to 1 of x^2 dx + integral from 1 to 2 of (x^2 - (4x - 4)) dx = [x^3/3] from 0 to 1 + [x^3/3 - 2x^2 + 4x] from 1 to 2 = 1/3 + (8/3 - 8 + 8) - (1/3 - 2 + 4) = 1/3 + 8/3 - 7/3 = 2/3 square units.