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Homework Help: Simple Integration by substitution

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Find by letting [tex]U^2=(4 + x^2) [/tex]the following [tex] \int_0^2\frac{x}{\sqrt{4 + x^2}}dx[/tex]?
    I can solve it by letting [tex]\mbox{x=2} tan(\theta)[/tex], But I want to be able to do it by substitution.
    3. The attempt at a solution
    [tex] \frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\[/tex] Therefore the integral is [tex] \int_{x=0}^{x=2}\frac{1}{u^\frac{1}{2}}du=[/tex]0.26757, it should be [tex] 2(\sqrt{2}-1)[/tex]. Can you tell me where I went wrong. Thanks for the help.
     
    Last edited: Feb 19, 2007
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  3. Feb 19, 2007 #2

    Dick

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    What happened to the x in the above expression for du? Did you just drop it?
     
  4. Feb 19, 2007 #3
    There is a mistake in the equation, it should be [tex]\int_0^2\frac{x}{\sqrt{4+x^2}}dx[/tex]. And the answer I then get =.5351. And thanks for the quick reply.
     
  5. Feb 19, 2007 #4

    Dick

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    Ah, ha. Then you want to evaluate the final integral between the u limits, not the x limits.
     
  6. Feb 19, 2007 #5

    Dick

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    And something is going awry here. A sqrt(u) would be the 4th root of 4+x^2. You are being sloppy.
     
  7. Feb 19, 2007 #6
    I know there is something awry in the calculation, maybe you can find what I did wrong. Since [tex]u^2=4+x^2 [/tex] that implies [tex] u=\sqrt{4+x^2}[/tex] and [tex] \frac{du}{dx}[/tex] expression follows.
     
    Last edited: Feb 19, 2007
  8. Feb 19, 2007 #7

    Dick

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    This shows du=(x*dx)/u. Not u^(1/2).
     
  9. Feb 19, 2007 #8
    First try substituting u=4+x²⇒du=2xdx⇒(1/2)du=xdx for the substitution part
    that should be a lot easier . you are still heading the right direction
    (1/2)∫(1/(u^{1/2}))du= √u
    I think you can take it from here
     
  10. Feb 19, 2007 #9

    Dick

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    He can also do it with his substitution. He just needs to be more careful of exactly what that is. :smile:
     
  11. Feb 20, 2007 #10
    Thanks for the help lads I at last got it out using my substitution.
     
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