The area (in sq. units) of the region described by $\left\{(x, y): y^2 \leq 2 x\right.$, and $\left.y \geq 4…
- $\frac{11}{32}$
- $\frac{8}{9}$
- $\frac{11}{12}$
- $\frac{9}{32}$
Solution

$\begin{aligned} & \text { Shaded area }=\int_{-\frac{1}{2}}^1\left(x_{\text {Right }}-x_{\text {Left }}\right) d y \\ & y^2=2 x \\ & y=4 x-1 \quad \text { Solve } \\ & y=1, y=-\frac{1}{2} \\ & \text { Shaded area }=\int_{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y \\ & =\left(\frac{1}{4}\left(\frac{y^2}{2}+y\right)-\frac{y^3}{6}\right)_{-\frac{1}{2}}^1=\frac{9}{32}\end{aligned}$
Asked in: JEE Main 2024 (04 Apr Shift 2)