The area (in sq. units) of the region described by $\left\{(x, y) / y^2 \leq 2 x\right.$ and $\left.y \geq(4 x-1)\right\}$ is

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Aicharya · Accepted Answer

Putting $x=\frac{y^2}{2}$ in $y=4 x-1$, we get $\begin{aligned} & y=4\left(\frac{y^2}{2}\right)-1 \Rightarrow 2 y^2-y-1=0 \\ & \Rightarrow(y-1)(2 y+1)=0 \\ & \Rightarrow y=1, \frac{-1}{2} \end{aligned}$ $\therefore \quad$ Required area $\begin{aligned} & =\int_{-1 / 2}^1\left(\frac{y+1}{4}\right) \mathrm{d} y-\int_{-1 / 2}^1 \frac{y^2}{2} \mathrm{~d} y \\ & =\frac{1}{4}\left[\frac{y^2}{2}+y\right]_{-1 / 2}^1-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-1 / 2}^1 \\ & =\frac{1}{4}\left[\left(\frac{1}{2}-\frac{1}{8}\right)+\left(1+\frac{1}{2}\right)\right]-\frac{1}{2}\left(\frac{1}{3}+\frac{1}{24}\right) \\ & =\frac{1}{4}\left(\frac{15}{8}\right)-\frac{1}{2}\left(\frac{9}{24}\right) \\ & =\frac{15}{32}-\frac{3}{16} \\ & =\frac{9}{32} \end{aligned}$

The area (in sq. units) of the region described by $\left\{(x, y) / y^2 \leq 2 x\right.$ and $\left.y \geq(4…

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