Question:
Find the area of the region bounded by the curve $y^{2}=4 x$ and the line $x=3$
Solution:
The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$.
The area OACO is symmetrical about x-axis.
∴ Area of OACO = 2 (Area of OAB)
Area $\mathrm{OACO}=2\left[\int_{0}^{3} y d x\right]$
$=2 \int_{0}^{3} 2 \sqrt{x} d x$
$=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3}$
$=\frac{8}{3}\left[(3)^{\frac{3}{2}}\right]$
$=8 \sqrt{3}$
Therefore, the required area is $8 \sqrt{3}$ units.