Question

Find the line y=mx through the origin, with positive slope, which together with the parabola y=x^2 encloses a region of area 4/3.

Answer 1

Find the point(s) of intersection

$$mx=x^2$$

$$x_1=0, x_2=m$$

$Point(0,0)$ and $Point(m, m^2)$

If $m>0$

$$Area=A=\displaystyle\int_{0}^{m}(mx-x^2)=[\dfrac{mx^2}{2}-\dfrac{x^3}{3}]\begin{matrix} m \\ 0 \end{matrix}$$

$$=\dfrac{m^3}{2}-\dfrac{m^3}{3}=\dfrac{m^3}{6} ({units}^2)$$

Substitute

$$\dfrac{m^3}{6}=\dfrac{4}{3}$$

$$m=2$$