Finding area bounded by two graphs of functions I know that someone posted this same question before here, because i found it, but i couldn't find the answer about this example that i was looking for, so i will post everything i did so far:
I have to find area bounded by graphics of functions 
$y_1=x\sqrt{4x-x^2}$
$y_2=\sqrt{4x-x^2}$
First, i found domains, the are the same for both of these functions $[0,4]$
Now that i have the domains i need to find intersection points, and there's one $x=1$ now , it means that graph of one function is above the graph of the other one up to on interval $[0,1]$ and after that they switch their roles on $[1,4]$
It turns out that  $y_1>y_2   (\forall x \in [0,1])$
and $y_2>y_1   (\forall x \in [1,4])$
So i have to find the following integral:
$P=\int_0^1 (\sqrt{4x-x^2} - x\sqrt{4x-x^2})dx  +  \int_1^4(x\sqrt{4x-x^2}-\sqrt{4x-x^2})dx$
This is equal to 
$P=\int_0^1 (1-x)\sqrt{4x-x^2}dx - \int_1^4 (1-x)\sqrt{4x-x^2}dx$
so i actually have to solve 
$\int (1-x)\sqrt{4x-x^2}dx = \int \sqrt{4x-x^2} - \int x\sqrt{4x-x^2}dx $
the first one can be easily solved using substitution $x=2sint$ and the second one can be solved using the same substitution except that we will have i little bit different solution.
It turns out that the solution of the first integral is
$2arcsin\frac{x-2}{2} + \frac{(x-2)\sqrt{4x-x^2}}{2}$
and for the second one:
$4arcsin\frac{x-2}{2} + (x-2)\sqrt{4x-x^2} + \frac{\sqrt{(4x-x^2)^3}}{3}$
Now when i insert the values i have:
$P=\int_0^1 \sqrt{4x-x^2}dx - \int_0^1 x\sqrt{4x-x^2}dx =[ 2arcsin\frac{1-2}{2} + \frac{(1-2)\sqrt{4*1-1^2}}{2} ] - [2arcsin\frac{0-2}{2} + \frac{(1-2)\sqrt{4*0-0^2}}{2}] - [2arcsin\frac{4-2}{2} + \frac{(4-2)\sqrt{4*4-4^2}}{2}] + [2arcsin\frac{1-2}{2} + \frac{(1-2)\sqrt{4*1-1^2}}{2}]$
In this large expression i have $arcsin(\frac{-1}{2})$ Now, i know that $arcsin(\frac{-1}{2}) = \frac{7\pi}{6} and \frac{11\pi}{6} $ my question is, which one i should use? In my solution, i used $\frac{7\pi}{6}$ And $\frac{11\pi}{6} = \frac{-\pi}{6}$ what about that?
And at the end, the most important thing, i got solution $P=\frac{-2\pi}{3}$ which is the negative value, but area can't be negative, what's wrong with this?
 A: Your integral is incorrect, you could check that on wolfram alpha.
For
$$\int_0^1 (1-x)\sqrt{4x-x^2}dx$$
Because $4x-x^2=4-(x-2)^2$
Let $x-2=2\sin t$
$$
\begin{align}
I
&=\int_{-\pi/2}^{-\pi/6} (-1-2\sin t)(2\cos t)(2\cos t)dt \\
&=-2\int_{-\pi/2}^{-\pi/6} (2\cos^2t -1+1)dt + 8\int_{-\pi/2}^{-\pi/3} \cos^2 t d(\cos t) \\
&=-\int_{-\pi/2}^{-\pi/6} \cos (2 t) d(2t) -2(-\pi/6+\pi/2) +\frac 83[ \cos^3 (-\pi/6)- \cos^3 (-\pi/2)]\\
&=-sin(-2\pi/6)+\sin(-2\pi/2) -\frac{2\pi}3 +\sqrt 3\\
&= \frac{3\sqrt 3}2-\frac {2\pi} 3
\end{align}
$$
You can use other methods to integrate it, a tip for you is that the range of arcsine is defined on $[-\pi/2,\pi/2]$, so you don't need to consider different values for arcsine. I will leave the second integral for you to evaluate
A: 
$P=\int_0^1 \sqrt{4x-x^2}dx - \int_0^1 x\sqrt{4x-x^2}dx $

You should have
$$P=\int_{0}^{1}\sqrt{4x-x^2}\ dx-\int_{0}^{1}x\sqrt{4x-x^2}\ dx\color{red}{-\left(\int_{1}^{4}\sqrt{4x-x^2}\ dx-\int_{1}^{4}x\sqrt{4x-x^2}\ dx\right)}$$
Also, $\arcsin(-1/2)=-\pi/6$ because the range of usual principal value of arcsine is $[-\pi/2,\pi/2]$.
I think you can continue from here.
