Integrating a strange square root function

[cutecarebear]

Homework Statement

Calculate the area under the curve for the following:
f(x) = √x*√(−32 − x)
and
g(x) = √(−x² − 32x)

Homework Equations

The Attempt at a Solution

I've been trying to do partial integration since the functions f(x)=g(x)area are mirrored, so they are equal area. I am thinking that since they have the same area, I only need to calculate one of the problems, but I just don't know how to approach it. I've tried using substitution, partial integration etc. I can't see any obvious trig identities. I'm just stuck. Thanks so much in advance!

 

[micromass]

You will eventually have to use trig identities. But it isn't immediately obvious which one. One hint is that you should complete the square of $$-x^2-32x$$. Then your integral will be in a more suitable form...

 

[HallsofIvy]

I'm not sure what you mean by "f(x) and g(x)" since those two functions are exactly the same. Of course, the function is only defined for x between -32 and 0 so the area under the curve will be
$$\int_{-32}^0 \sqrt{-32x- x^2}dx$$

If you make the change of variable u= -x that gives
$$\int_0^{32}\sqrt{u^2+ 32u}dx$$
which isn't really any simpler but may look nicer.

As micromass suggests, complete the square: $u^2+ 32u= u^2+ 32u+ 256- 256= (u+ 16)^2- 256$ and the substitution v= u+ 16 further reduces the integral to
[tex]\int_{16}^{48}\sqrt{v^2- 256}dv[/itex]<br /> <br /> Remembering that $sin^2+ cos^2= 1$, $tan^2+ 1= sec^2$ and $sec^2- 1= tan^2$. That is, the substitution $v= 16 sec(\theta)$ will get rid of the square root:<br /> $$\sqrt{v^2- 256}= \sqrt{256sec^(\theta)- 256}= 16\sqrt{sec^2(\theta)- 1}= 16tan(\theta)$$.[/tex]