What is the value of $a$ so that this condition holds? Let $f(x) \colon= x-x^2$, $g(x) \colon= ax$. Determine the value of $a$ so that the region above the graph of $g$ and below the graph of $f$ has area equal to $9/2$. 
Here $f(x) - g(x) = (1-a)x - x^2 = x((1-a) - x) = 0$ if and only if $x = 0$ or $x = 1-a$. 
So the required area is $$ \int_{0}^{1-a} ( (1-a)x - x^2 ) \ dx = \frac{(1-a)^2}{2} - \frac{(1-a)^3}{3} = \frac{(1-a)^2}{6} ( 3 - 2(1-a) ) =  \frac{(1-a)^2}{6} (2a + 1) =  \frac{(1-a)^2 (2a+1) }{6} =  \frac{(a^2 -2a +1 ) (2a+1)}{6} = \frac{2a^3 - 3a^2 + 1}{6} = \frac{9}{2},  $$ 
and this last equation simplifies to $$ 2a^3 - 3a^2 - 26 = 0.$$ 
Is the process so far correct? And if so, then how to solve this last equation for $a$? 

You should have gotten $\displaystyle\int_{0}^{1-a}[(1-a)x-x^2]\,dx = \left[\dfrac{1-a}{2}x^2-\dfrac{1}{3}x^3\right]_{0}^{1-a} = \dfrac{(1-a)^3}{6}$. 
Then, you have $\dfrac{(1-a)^3}{6} = \dfrac{9}{2} \leadsto (1-a)^3 = 27$ which should be easy to solve. 
