Question

Find the area of the region included between $y^2=9xandy=x$.

Answer 1

We have, $y^2=9xandy=x$

$\Rightarrow x^2=9x$

$\Rightarrow x^2-9x=0$

$\Rightarrow x(x-9)=0$

$\Rightarrow x=0,9$

$\therefore$ Area of shaded region, $A={\int }_0^9(\sqrt{9x}-x)dx={\int }_0^{9}3x^{\frac{1}{2}}dx-{\int }_0^{9}xdx={\left[3.\frac{x^{\frac{3}{2}}}{3}.2\right]}_0^9-{\left[\frac{x^2}{2}\right]}_0^9[=\frac{{3.3}^{\frac{3}{2}\times 2}}{3}.2-0]-[\frac{81}{2}-0]=54-\frac{81}{2}=\frac{108-81}{2}=\frac{27}{2}sq.units$