The area (in sq.

Question

The area (in sq. units) of the region $\{(x,y)\in\mathbb R^2 \mid 4x^2\le y\le8x+12\}$ is:

Accepted Answer

$4x^2 = 8x+12 \Rightarrow x^2-2x-3=0 \Rightarrow (x+1)(x-3)=0 \Rightarrow x=-1,3$.

$$A = \int_{-1}^3\left(8x+12-4x^2\right)dx = \left[4x^2+12x-\frac{4x^3}{3}\right]_{-1}^3 = (36+36-36)-\left(4-12+\frac43\right)$$

$$= 36 - \left(-8+\frac43\right) = 36+8-\frac43 = \frac{128}{3}$$

Solutions · 1