Question

Find the area of region bounded.
$y^2=4x$ and the line $x=3$

Answer 1

Let $AB$ represent the line $x=3$ and $AOB$ represent the curve $y^2=4x$

Area of $AOBC=2\times$Area of $AOC$

$=2\times {\int }_0^{3}y.dx$

We know that $y^2=4\Rightarrow y=\pm 2\sqrt{x}$

As $AOC$ is in first quadrant

$\therefore y=2\sqrt{x}$

$\therefore$Area of $AOBC=2\times {\int }_0^{3}y.dx$

$=2\times {\int }_0^{3}2\sqrt{x}.dx$

$=4\times {\int }_0^3\sqrt{x}.dx$

$=4\times {\int }_0^3{x}^{\frac{1}{2}}.dx$

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

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

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

$=\frac{8}{3}[3^{\frac{3}{2}}-0]$

$=\frac{8}{3}\times 3\sqrt{3}$

$=8\sqrt{3}$ sq.units

$\therefore$ Required area$=8\sqrt{3}$ sq.units