Mathematics - Area Under Curves Question with Solution | TestHub

The equation of curve is $y=x-x^2$ $\begin{array}{l}\Rightarrow x^2-x=y\\ \Rightarrow {\left(x-\frac{1}{2}\right)}^2=-\left(y-\frac{1}{4}\right)\end{array}$ which is a parabola whose vertex is $\left(\frac{1}{2},\frac{1}{4}\right)$
Hence, finding the point of intersection of the curve and the line, $x-x^2=mx\Rightarrow x(1-x-m)=0$ i.e., $x=0$ or $x=1-m$ $\therefore \frac{9}{2}={\int }_0^{1-\mathrm{m}}\left(\mathrm{x}-{\mathrm{x}}^2-\mathrm{mx}\right)\mathrm{dx}$ $={\left\{\frac{x^2}{2}-\frac{x^3}{3}-m\frac{x^2}{2}\right\}}_0^{1-m}$ $=(1-\mathrm{m})\frac{(1-\mathrm{m}{)}^2}{2}-\frac{(1-\mathrm{m}{)}^3}{3}=\frac{(1-\mathrm{m}{)}^3}{6}$ $\therefore (1-\mathrm{m}{)}^3=\frac{6\times 9}{2}=27$ $\Rightarrow 1-\mathrm{m}=(27{)}^{1/3}=3$ $\Rightarrow \mathrm{m}=-2$ Also, $(1-\mathrm{m}{)}^3-(3{)}^3=0$ $\therefore (1-\mathrm{m}{)}^3=3^3\Rightarrow 1-\mathrm{m}=3$ or $\mathrm{m}=-2$