The area (in square units) of the region bounded by $y^2 = 2x$ and $y =4x-1$, is

$\frac{9}{32}$

The shaded region in Fig is the region bounded by $y^2 = 2x$ and $y = 4x-1$. We slice this region into horizontal strips. The approximating rectangle shown in Fig has length = $(x_2-x_1)$, width = Δy and area = $(x_2-x_1) Δy$. As it can move vertically between A and B.

So, required area A of the given region is given by

$∴A=\int\limits_{-1/2}^{1}(x_2-x_1)dy$

$⇒A=\int\limits_{-1/2}^{1}\left(\frac{y+1}{4}-\frac{y^2}{2}\right)dy$ [∵$P(x_1,y)$ and $Q(x_2,y)$ lie on $y^2=2x\And y=4x-1$ respectively. $∴y=4x_1\,and\,y^2=2x_2$]

$⇒A=\left[\frac{(y+1)^2}{8}-\frac{y^3}{6}\right]_{-1/2}^{1}=\frac{1}{3}-\left(\frac{1}{32}+\frac{1}{48}\right)=\frac{9}{32}$ sq. units