Let A = (x,y) : y^24x, y-2x-4.

Question

Let $A = \{(x,y) : y^2\le4x,\ y-2x\ge-4\}$. The area (in square units) of the region $A$ is:

Accepted Answer

Write $x\le\dfrac{y+4}{2}$ and $x\ge\dfrac{y^2}{4}$; these meet where $\dfrac{y^2}{4}=\dfrac{y+4}{2} \Rightarrow y^2-2y-8=0 \Rightarrow y=4,-2$.

$$\text{Area} = \int_{-2}^4\left[\frac{y+4}{2}-\frac{y^2}{4}\right]dy = \frac12\left[\frac{y^2}{2}+4y\right]_{-2}^4 - \frac14\left[\frac{y^3}{3}\right]_{-2}^4$$

$$= \frac12\bigl[(8+16)-(2-8)\bigr] - \frac14\left(\frac{64}{3}+\frac83\right) = 9$$

Solutions · 1