Question

Find the area of the region

Y=x*sqrt((4-x)/(4+x)), y=0, x=4

Answer 1

just plug in the values :) given for "x" and "y" in the equation

Answer 2

It's integration

Answer 3

\[y=x\sqrt{\frac{4-x}{4+x}}\]
Here's a link with a pic of the region: http://www.wolframalpha.com/input/?i=Plot%5Bx*Sqrt%5B%284-x%29%2F%284%2Bx%29%5D%2C%7Bx%2C-1%2C4%7D%5D

\[A=\int_0^4 x\sqrt{\frac{4-x}{4+x}}~dx\]
I'd start off with a substitution: \(u=4+x~\iff~x=u-4\), so \(du=dx\):
\[A=\int_4^8 (u-4)\sqrt{\frac{4-(u-4)}{u}}~du\\
A=\int_4^8 (u-4)\sqrt{\frac{8-u}{u}}~du
\]
Then another substitution: \(t=\dfrac{8-u}{u}~\iff~u=\dfrac{8}{t+1}\),
so \(dt=-\dfrac{8}{u^2}~du~\iff~du=-\dfrac{\left(\frac{8}{t+1}\right)^2}{8}~dt=-\dfrac{8}{(t+1)^2}~dt\).

So you have \(u-4=\dfrac{8}{t+1}-4=\dfrac{4-4t}{t+1}\), and the integral becomes
\[A=\int_1^0 \dfrac{4-4t}{t+1}\sqrt{t}~\left(-\dfrac{8}{(t+1)^2}~dt\right)\\
A=32\int_0^1 \dfrac{1-t}{(t+1)^3}\sqrt{t}~dt\]
Yet another substitution: \(s=\sqrt{t}~\iff~s^2=t\), so \(ds=\dfrac{1}{2\sqrt{t}}~dt~\iff~dt=2s~ds\).
\[A=32\int_0^1 \dfrac{1-s^2}{(s^2+1)^3}s~(2s~ds)\\
A=64\int_0^1 \dfrac{(1-s^2)s^2}{(s^2+1)^3}~ds\]
Break up into partial fractions. The work from there gets easier.