JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
The area of the region $S=\{(x,y):3x^{2}\le4y\le6x+24\}$ is $\_\_\_\_\_\_$ .
Application of Integrals
Answer$27$
Solution
$3x^2\le4y\le6x+24$: $\frac{3x^2}{4}\le y\le\frac{3x}{2}+6$. Intersection: $x^2-2x-8=0\Rightarrow x=-2,4$. Area $=\int_{-2}^4\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx=\left[\frac{3x^2}{4}+6x-\frac{x^3}{4}\right]_{-2}^4=(12+24-16)-(3-12+2)=20+7=27$.