# Integrating a strange square root function

## Homework Statement

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

## 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!

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
Homework Helper
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
$$\int_{16}^{48}\sqrt{v^2- 256}dv[/itex] Remembering that $sin^2+ cos^2= 1$, $tan^2+ 1= sec^2$ and $sec^2- 1= tan^2$. That is, the substitution $v= 16 sec(\theta)$ will get rid of the square root: [tex]\sqrt{v^2- 256}= \sqrt{256sec^(\theta)- 256}= 16\sqrt{sec^2(\theta)- 1}= 16tan(\theta)$$.