Question

For which value of m the area bounded by the line y=MX and Y=X-X^2 is 9÷2

Answer 1

Start by finding the intersection points :
$mx = x-x^2\\x(m-1+x)=0\\x=0,~1-m$

Setup the area integral, set it equal to $\frac{9}{2}$ and solve $m$
$\int\limits_0^{1-m}x-x^2-mx\,dx=\frac{9}{2}$

$\int\limits_0^{1-m}(1-m)x-x^2\,dx=\frac{9}{2}$

$(1-m)\frac{x^2}{2}-\frac{x^3}{3}\Bigg|_{0}^{1-m}=\frac{9}{2}$

$(1-m)\frac{(1-m)^2}{2}-\frac{(1-m)^3}{3}=\frac{9}{2}$

$\frac{(1-m)^3}{6}=\frac{9}{2}$

$(1-m)^3=27$

$1-m=3$

$m=\boxed{-2}$