Find the area bounded by the curve


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


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

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

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




Similarly, Area OACO = Area OLAC – Area OLAO

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

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





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

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