Mathematics - Area Under Curves Question with Solution | TestHub

Given, the equation of line, $y=mx$

And the line $x=0, y=0, x=\frac{3}{2}$

The intersection point $\left(\frac{3}{2},0\right)$

Area of triangle is half the total area 

$\frac{1}{2}{\int }_0^{3/2}\left(1+4x-x^2\right)dx={\int }_0^{3/2}mxdx$

$\frac{1}{2}{\left[x+2x^2-\frac{x^3}{3}\right]}_0^{3/2}=\frac{m}{2}{\left[x^2\right]}_0^{3/2}$

$\left(\frac{3}{2}+\frac{9}{2}-\frac{9}{8}\right)=\frac{9m}{4}$

$\frac{39}{8}=\frac{9m}{4}$

$m=\frac{39}{18}$

Then, $12m$

$12 m=\frac{39}{18}\times 12$

$12m=26$