The area (in sq. units) of the region $\mathrm{A}=\left\{(x, y) / \frac{y^2}{2} \leq x \leq y+4\right\}$

Question

Aicharya · Accepted Answer

Given that $\frac{y^2}{2} \leq x \leq y+4$ $\begin{array}{ll} 	herefore \quad & x=\frac{y^2}{2} 	ext { and } x=y+4 \ & \frac{y^2}{2}=y+4^2 \ 	herefore \quad & y^2-2 y-8=0 \ 	herefore \quad & y=4 	ext { or }-2 \ & \Rightarrow x=8 	ext { or } 2 \end{array}$ $	herefore \quad \mathrm{A}=\int_{-2}^4\left(y+4-\frac{y^2}{2}ight) \mathrm{d} y$ $\begin{aligned} & 	herefore \quad \mathrm{A}=\left[\frac{y^2}{2}+4 y-\frac{y^3}{6}ight]_{-2}^4 \ & 	herefore \quad A=\left(8+16-\frac{64}{6}ight)-\left(2-8+\frac{8}{6}ight) \ & 	herefore \quad \mathrm{A}=18 \ & \end{aligned}$

The area (in sq. units) of the region $\mathrm{A}=\left\{(x, y) / \frac{y^2}{2} \leq x \leq y+4\right\}$

Solution