Question

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

Answer 1

The required area is twice the area bounded by the curve $y=2\sqrt{a}\sqrt{x},x=0,x=a$ (latus rectum) and $x$-axis. [Since it is symmetrical about $x$ - axis].
Required area $=2{\int }_0^{a}ydx$
$=2{\int }_0^{a}2\sqrt{a}.\sqrt{x}dx$

$=4\sqrt{a}.{\left[\frac{x^{3/2}}{3/2}\right]}_0^a$
$=\frac{8\sqrt{a}}{3}[a^{3/2}-0]$

$=\frac{8a^2}{3}$ sq. units