<s>[INST] 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 ?
 [/INST] Alternatively, note that $Q=2P \Rightarrow P+Q=3P$. Make up the equation:
\begin{align}\int _{ 0 }^{ 2a } \left( 3x-{ x }^{ 2 } \right) \text dx  &=3\int _{ 0 }^{ a } \left( 3x-{ x }^{ 2 } \right) \text dx \\
\left.\left(\frac{3x^2}{2}-\frac{x^3}{3}\right)\right|_{0}^{2a}&=3\left.\left(\frac{3x^2}{2}-\frac{x^3}{3}\right)\right|_{0}^{a}\\
\left(\frac{12a^2}{2}-\frac{8a^3}{3}\right)&=3\left(\frac{3a^2}{2}-\frac{a^3}{3}\right)\\
\frac{36a^2-16a^3}{6}&=\frac{27a^2-6a^3}{6}\\
10a^3-9a^2&=0\\
a^2(10a-9)&=0 \stackrel{a\ne0}\Rightarrow 10a-9=0 \Rightarrow a=\fbox{$\frac{9}{10}$}\end{align}
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