Find the area of the region bounded by the parabola y 2 = 4ax and its latus rectum x = a (full region, both above and below the x-axis).

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Find the area of the region bounded by the parabola y² = 4ax and its latus rectum x = a (full region, both above and below the x-axis).

Answer: (8/3)a².

A (8/3)a²
B (16/3)a²
C (4/3)a²
D 4a²

Correct answer: A. (8/3)a²

Explanation: By symmetry, total area = 2 times the upper-half area. Upper-half area = integral from 0 to a of 2sqrt(ax) dx = 2sqrt(a)[(2/3)x^3/2] from 0 to a = 2sqrt(a)(2/3)a^3/2 = (4/3)a^2. Doubling for both halves gives 2 times (4/3)a² = (8/3)a^2.

The definite integral ∫ₐᵇf(x)dx computes the exact area of the shaded region bounded by the curve, the x-axis, and the vertical lines x=a and x=b - the same idea behind the Riemann sum, but evaluated exactly rather than approximated by rectangles.

Concept context

Using definite integrals to compute the area under a curve, the area between two curves, and the areas enclosed by standard curves like circles, parabolas, and ellipses.

Read the full Application of Integrals notes →

Find the area of the region bounded by the parabola y2 = 4ax and its latus rectum x = a (full region, both above and below the x-axis).

Concept context

More Application of Integrals questions

Find the area of the region bounded by the parabola y² = 4ax and its latus rectum x = a (full region, both above and below the x-axis).