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

Let us find the points intersections of $y^2=2x$ and $y=4x-1$

$\Rightarrow {\left(4x-1\right)}^2=2x$

$\Rightarrow 16x^2-10x+1=0$

$\Rightarrow \left(8x-1\right)\left(2x-1\right)=0$

$\mathrm{\Rightarrow } x=\frac{1}{2},\frac{1}{8}$

$x=\frac{1}{2} \Rightarrow y=4\left(\frac{1}{2}\right)-1=1 \Rightarrow A\left(\frac{1}{2},1\right)$

$x=\frac{1}{8} \Rightarrow y=4\left(\frac{1}{8}\right)-1=\frac{-1}{2}\Rightarrow B\left(\frac{1}{8},\frac{-1}{2}\right)$

$A_1={\int }_0^{\frac{1}{8}}\sqrt{2x}dx + \frac{1}{2}\times \frac{1}{8}\times \frac{1}{2}$

$A_1=\sqrt{2}{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^{\frac{1}{8}} + \frac{1}{32}$

$A_1=\frac{1}{24} + \frac{1}{32} \mathrm{sq}.\mathrm{units}$

$A_2={\int }_0^{\frac{1}{2}}\sqrt{2x}dx - \frac{1}{2}\times 1\times \frac{1}{4}$

$A_2=\sqrt{2}{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^{\frac{1}{2}} - \frac{1}{8}$

$A_2=\frac{5}{24} \mathrm{sq}.\mathrm{units}$

Therefore, the required area is, 

$A=\frac{1}{24}+\frac{1}{32}+\frac{5}{24}=\frac{9}{32} \mathrm{sq}.\mathrm{units}$