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Integrating a strange square root function

  • #1

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!
 

Answers and Replies

  • #2
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3,281
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 [tex]-x^2-32x[/tex]. Then your integral will be in a more suitable form...
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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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
[tex]\int_{-32}^0 \sqrt{-32x- x^2}dx[/tex]

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

As micromass suggests, complete the square: [itex]u^2+ 32u= u^2+ 32u+ 256- 256= (u+ 16)^2- 256[/itex] and the substitution v= u+ 16 further reduces the integral to
[tex]\int_{16}^{48}\sqrt{v^2- 256}dv[/itex]

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

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