Mathematics - Area Under Curves Question with Solution | TestHub

$0\le 9x\le y^2\&y\ge 3x-6$

$\begin{array}{rl}& A=\text{ Required Area }=\left[{\int }_0^1(-3\sqrt{x})dx-{\int }_0^1(3x-6)dx\right]\\ & A=-{3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)|}_0^1-{\left(\frac{3x^2}{2}-6x\right)|}_0^1\\ & A=-2[1-0]\left[\frac{3}{2}-6\right]\\ & A=-2-\frac{3}{2}+6=\frac{5}{2}\text{ Sq. unit }\\ & \therefore 6A=6\times \frac{5}{2}=15\end{array}$