The area (in sq.

Question

The area (in sq. units) of the region $\{(x,y)\in\mathbb R^2 : x^2\le y \le |3-2x|\}$ is:

Accepted Answer

For $x\le3/2$, $|3-2x|=3-2x$; the parabola $y=x^2$ meets $y=-2x+3$ where $x^2+2x-3=0 \Rightarrow x=-3,1$. Taking the branch in the first quadrant region (up to $x=1$):

$$\text{Area} = \int_{-3}^{1}\left(3-2x-x^2\right)dx = \left[3x-x^2-\frac{x^3}{3}\right]_{-3}^1 = \left(3-1-\frac13\right)-\left(-9-9+9\right) = \frac{32}{3}$$

Solutions · 1