Q. The area of the region $\{(x,y):x^2+4x+2\le y\le|x+2|\}$ is equal to
A) $7$
B) $5$
C) $\frac{24}5$
D) $\frac{20}3$
Explanation
[Solution by Dr. Abubacker] Put $u=x+2$: the region is $u^2-2\le y\le|u|$. The curves meet where $u^2-|u|-2=0$, i.e. $|u|=2$. Area $=\int_{-2}^{2}\left(|u|-u^2+2\right)du=2\int_0^2(u-u^2+2)\,du=2\left(2-\frac83+4\right)=\frac{20}3$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.