Question

Find the area bounded by y² = 4x and y + 2x = 12.

Answer 1

Let us express both functions as $x = x(y)$:

$f_1(y) = \frac{y^2}{4}, f_2(y) = 6 - \frac{y}{2}$.

The two intersect at $y = -6, y = 4$, hence the area between them is:

$S = \int_{-6}^4 [(6-\frac{y}{2}) - \frac{y^2}{4}] dy = (6 y - \frac{y^2}{4} - \frac{y^3}{12})|_{-6}^4 = \frac{125}{3}$