Question

If $y=mx$ bisects the area enclosed by the lines $x=0,y=0,x=\frac{3}{2}$ and the curve $y=1+4x-x^2$ then $6m$ is equal to

Answer 1

${\int }_0^{3/2}(1+4x-x^2)dx=2{\int }_0^{3/2}mxdx$
$\Rightarrow {|x+2x^2-x^3/3|}_0^{3/2}={|m{x}^2|}_0^{3/2}$
$\Rightarrow 6m=13$