Question

The area (in sq. units) of the region $\left\{\right(x,y)\in R^2|4x^2\le y\le 8x+12\}$ is :

• A. $\frac{125}{3}$
• B. $\frac{128}{3}$
• C. $\frac{124}{3}$
• D. $\frac{127}{3}$

Answer 1

The correct option is B $\frac{128}{3}$


For point of intersection,
$4x^2=8x+12$
$\Rightarrow x^2-2x-3=0$
$\Rightarrow x=3,-1$
Area bounded is given by
$A=\underset{-1}{\overset{3}{\int }}(8x+12-4x^2)dx$
$\Rightarrow A={[\frac{8x^2}{2}+12x-\frac{4x^3}{3}]}_{-1}^3$
$\Rightarrow A=(36+36-36)-(4-12+\frac{4}{3})$
$\Rightarrow A=44-\frac{4}{3}=\frac{128}{3}$