Question

The area (in sq. units) of the region $\left\{\left(x,y\right):y^2\le 2xandy\ge 4x-1\right\}$ is :

• A. $\frac{4}{3}$
• B. $\frac{7}{3}$
• C. $\frac{4\sqrt{2}}{3}$
• D. $\frac{9}{32}$
  

Answer 1

The correct option is D $\frac{9}{32}$

$\begin{array}{c}{y}^2\le 2x\\ and,\\ y\ge 4x-1\\ Bysolving{y}^2=2x\\ and,y=4x-1\\ \therefore y=1,\frac{-1}{2}\\ A={\int }_{\frac{-1}{2}}^1\left[\left(\frac{y+1}{4}\right)-\left(\frac{y^2}{2}\right)\right]dy=\frac{9}{32}.\end{array}$