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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 $a$ .

graph

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 ?

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

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Since $a\neq 0$ (as this would be absurd), we can solve your equation as follows

\begin{align}9a^2−2a^3&=18a^2−4a^3\\ 9-2a&=18-4a\tag{divide through by $a$}\\ 2a&=9\\ a&=\frac 92\end{align}

However, as noted in other answers, this is incorrect.

Your error is somewhere in your calculation of the area of $Q$, you should get $$\int_a^{2a}(3x-x^2)\text dx =\frac{27a^2-14a^3}{6}$$

This then gives you \begin{align}Q&=2P\\\frac{27a^2-14a^3}{6}&=2\times \frac{9a^2-2a^3}{6}\\ 27a^2-14a^3&=2(9a^2-2a^3)\\ 27a^2-14a^3&=18a^2-4a^3\\ 27-14a&=18-4a\tag{divide through by $a$}\\ 9&=10a\\ a&=\frac 9{10}\end{align}

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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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you have that

$2\int _0^a\:\left(3x-x^2\right)dx\:=\int _a^{2a}\:\left(3x-x^2\right)dx\:$

so

$\left(-3\:a^2\right)/2\:+\:\left(5\:a^3\right)/3\:=\:0$

the solution to this equation are:

$a=0,\:a=\frac{9}{10}$ which is your answer

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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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  • $\begingroup$ Possibly it's simpler to use \boxed{ ... } than \fbox{ $ ... $ } $\endgroup$ – CiaPan Jun 22 '17 at 14:08

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