Question

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

Answer 1

Area bound by $y^2=4x$ and $x=3$

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

$x=3,y^2=12$

$y=\pm 2\sqrt{3}$

$y^2=4x$ is symmetric wrt to x axis area of $OAB=2OAC$

$A=\int ydx$

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

$=\frac{2}{\frac{3}{2}}{x}^{\frac{3}{2}}$

$△OAC=\frac{4}{3}\times {\left(27\right)}^{\frac{1}{2}}=4\sqrt{3}$

$△OAB=2△OAC=8\sqrt{3}$