Area under the graph - integration 
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .


Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?

$$\int _{ a }^{ 2a }{ \left( 3x-{ x }^{ 2 } \right) dx } =2\int _{ 0 }^{ a }{ \left( 3x-{ x }^{ 2 } \right) dx } \\ \frac { 36{ a }^{ 2 }-16{ a }^{ 3 } }{ 6 } -\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ \frac { 27{ a }^{ 2 }-14{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ 27{ a }^{ 2 }-14{ a }^{ 3 }=18{ a }^{ 2 }-4{ a }^{ 3 }\\ 9{ a }^{ 2 }-10{ a }^{ 3 }=0\\$$  clearly $  a\neq 0$ so the answer is  $$   \color{red}{a=\frac { 9 }{ 10 }}   $$
