Find the area bounded by the curve

Solution:

The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area $\mathrm{OBAO}$.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point $A$ are $\left(-1, \frac{1}{4}\right)$.

Coordinates of point $\mathrm{B}$ are $(2,1)$.

We draw AL and BM perpendicular to $x$-axis.

It can be observed that,

Area $\mathrm{OBAO}=$ Area $\mathrm{OBCO}+$ Area $\mathrm{OACO}$

 

Then, Area OBCO = Area OMBC - Area OMBO

$=\int_{0}^{2} \frac{x+2}{4} d x-\int_{0}^{2} \frac{x^{2}}{4} d x$

$=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{0}^{2}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{0}^{2}$

$=\frac{1}{4}[2+4]-\frac{1}{4}\left[\frac{8}{3}\right]$

$=\frac{3}{2}-\frac{2}{3}$

$=\frac{5}{6}$

Similarly, Area OACO = Area OLAC – Area OLAO

$=\int_{-1}^{0} \frac{x+2}{4} d x-\int_{-1}^{0} \frac{x^{2}}{4} d x$

$=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{0}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{-1}^{0}$

$=-\frac{1}{4}\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\left[-\frac{1}{4}\left(\frac{(-1)^{3}}{3}\right)\right]$

$=-\frac{1}{4}\left[\frac{1}{2}-2\right]-\frac{1}{12}$

$=\frac{1}{2}-\frac{1}{8}-\frac{1}{12}$

$=\frac{7}{24}$

Therefore, required area $=\left(\frac{5}{6}+\frac{7}{24}\right)=\frac{9}{8}$ units