If the area of the region bounded by the line y = mx and the curve x^2= y is 32/3 sq. units, then find the positive value of m, using integration.

\(\displaystyle x\)-coordinates of points of intersection are \(\displaystyle 0, m\). \[\begin{aligned} & \text { According to question }=\int_{0}^{m}\left(m x-x^{2}\right) d x=\frac{3}{3} \\ & \Rightarrow m \frac{x^{2}}{2}-\left.\frac{x^{3}}{3}\right|_{0} ^{m}=\frac{32}{3} \\ & \frac{m^{3}}{6}=\frac{32}{3} \Rightarrow m^{3}=64 \\ & \Rightarrow m=4 \end{aligned} \]