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Homework Help: Definite Integral (Square Root)

  1. Sep 4, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP184619c56eg84971107d00001e292h4340heh8g4?MSPStoreType=image/gif&s=35&w=135&h=53 [Broken]

    It is a definite integral between x=0 and x=20.

    I forget how to integrate with a square root. Do you just change it to a power of a half and then integrate it?

    Been a while since I've done this,

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 4, 2010 #2

    danago

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    I would use a trigonometric substitution for this one.

    Let [tex]x=20 sin(\theta)[/tex]
     
  4. Sep 4, 2010 #3
    Not quite understanding. Since this is a definite integral, if I let x=20sin(theta), do I then sub in 0 and 20 as theta?

    There doesn't seem to be any trigonmetric substitution in the text book? Is there anyway that formula could be arranged to fit with the short table of standard integrals?
     
    Last edited: Sep 4, 2010
  5. Sep 4, 2010 #4

    danago

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    Gold Member

    Well if [tex]x=20 sin(\theta)[/tex], then [tex]\theta = sin^{-1}(\frac{x}{20})[/tex]. x=0 is transformed to [tex]\theta=0[/tex] and x=20 is transformed to [tex]\theta=\frac{\pi}{2}[/tex].

    [tex]\int\limits_0^{20} {\sqrt {1 - \frac{{{x^2}}}{{400}}} \;dx = \int\limits_0^{\pi /2} {20\cos \theta \sqrt {1 - {{\sin }^2}\theta } \;\;d\theta } } [/tex]

    Can you work from there?

    Im sure that if you looked at a table of standard integrals you would find one that can be used to solve this.
     
  6. Sep 4, 2010 #5
    I can get the answer through trigonometric substitution, the problem is it isn't in my textbook so therefore not part of the syllabus.

    I can integrate it fine by myself if I know the rule. All I've got is some basic substitution stuff (chain rule) and standard integrals, but I can't find a rule.

    I thought if I rearranged the equation from:
    Sqrt(16^2 (1 - x^2/20^2))

    To:
    4/5 sqrt(400-x^2)

    You could integrate that to:

    (x Sqrt[a^2 - x^2])/2 - (a^2 ArcCos[x/a])/2

    where a=20 and x=20 or 0.

    But that doesn't seem to work, especially since I know the answer to this.

    All I need is to get the working out, so just the formula.

    (P.S How does Latex work on this)

    Nevermind, solved.
     
    Last edited: Sep 4, 2010
  7. Sep 5, 2010 #6

    danago

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    For information about the Latex used on this forum, check out this link:

    https://www.physicsforums.com/showthread.php?t=386951 [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Sep 5, 2010 #7

    HallsofIvy

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    The limits of integration, in the original integral, are in terms of the variable in that integral: x= 0 and x= 20. If you then use [itex]x= 20 sin(\theta)[/itex] then you have [itex]x= 0= 20 sin(\theta)[/itex], then [itex]sin(\theta)= 0[/itex] so [itex]\theta= 0[/itex] and [itex]x= 20= 20sin(\theta)[/itex], then [itex]sin(\theta)= 1[/itex] so [itex]\theta= \pi/2[/itex].
     
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