<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] Area of $Q$ certainly is not $$\frac{9a^2-2a^3}{6}$$
It is:
$$Q = \int_a^{2a}\ (3x-x^2)dx=\left .\frac{9x^2-2x^3}6\right |_{x=a}^{x=2a}$$
$$= \frac{9(2a)^2-2(2a)^3}6 - \frac{9a^2-2a^3}6 $$
$$= \frac {36a^2-16a^3}6 - \frac{9a^2-2a^3}6$$
$$= \frac {27a^2-14a^3}6$$
so $Q=2P$ yields
$$27a^2-14a^3 = 2(9a^2-2a^3)$$
$$27a^2-14a^3 = 18a^2-4a^3$$
which others solved already.
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