Question

Find the area bounded by the curve $x^2=4y$ and the line $x=4y-2$.

Answer 1

The area bounded by the curve, $x^2=4y$, and line, $x=4y-2$, is represented by the shaded area $OBAO$.
Let $A$ and $B$ be the points of intersection of the line and parabola.
Coordinates of point $A\text{are}(-1,\frac{1}{4})$
Coordinates of point $B$ are $(2,1)$.
We draw $AL$ and $BM$ perpendicular to x-axis.

It can be observed that,

$\text{Area}OBAO=\text{Area}OBCO+\text{Area}OACO$ ........... (1)

Then, $\text{Area}OBCO=\text{Area}OMBC-\text{Area}OMBO$

$={\int }_0^2\frac{x+2}{4}dx-{\int }_0^2\frac{x^2}{4}dx$

$=\frac{1}{4}{[\frac{x^2}{2}+2x]}_0^2-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^2$

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

$=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}$

Similarly, Area $OACO$ $=$ Area $OLAC$ - Area $OLAO$

$={\int }_{-1}^0\frac{x+2}{4}dx-{\int }_{-1}^0\frac{x^2}{4}dx$

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

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

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

$=\frac{1}{2}-\frac{1}{8}-\frac{1}{12}=\frac{7}{24}$
Therefore, required area $=(\frac{5}{6}+\frac{7}{24})=\frac{9}{8}$ sq. units.