Definite Integral

  1. Jul 22, 2008 #1

    LHC

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    The question in my textbook was:

    [tex]\int_{0}^{2} x^2 \sqrt{4-x^2} dx[/tex]

    I decided to just leave out the lower and upper limits for now, and just solve [tex]\int x^2 \sqrt{4-x^2} dx[/tex].

    (It's a bit long, but I assure you I did the work.) Upon making the substitution of [tex]x = 2 \sin \theta[/tex], I got it down to:

    [tex]\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \sin^3 \theta + C[/tex]

    Now, I'm transforming it back in terms of x, so [tex]\sin \theta = \frac{x}{2}[/tex]

    So, I thought it would make this whole thing:

    [tex]\int x^2 \sqrt{4-x^2} dx = \frac{8}{3} \times \frac{x^3}{8}[/tex], so

    [tex]\int x^2 \sqrt{4-x^2} dx = \frac{x^3}{3}[/tex]

    If I finish up the problem by using the limits of 0 and 2, I get:

    [tex]\int_{0}^{2} x^2 \sqrt{4-x^2} dx = \frac{8}{3}[/tex]


    But the answer I got from the calculator was around 3.14 (not pi, though). Could someone please tell me where I went wrong? Thanks
     
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  3. Jul 22, 2008 #2

    Dick

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    Well, something went wrong in the long part you aren't telling us about. Because the derivative of x^3/3 is DEFINITELY not x^2*sqrt(4-x^2).
     
  4. Jul 22, 2008 #3

    nicksauce

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    Also, FYI the integral should be exactly pi (so says maple).
     
  5. Jul 22, 2008 #4

    LHC

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    Well, I think it's some misunderstanding on my part from the conversion between x and theta.

    I used the Mathematica online integrator to verify and its answer was:

    [tex]\frac{8}{3} \sqrt{\cos^2 x} \sin^2 x \tan x[/tex]...which, when I simplify (and perhaps this is the part where I'm wrong), I just end up with [tex]\frac{8}{3} \sin^3 \theta [/tex]... which is where I started off =P
     
  6. Jul 22, 2008 #5

    Dick

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    What did you put into the integrator??
     
  7. Jul 22, 2008 #6

    LHC

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    (2sinx)^2*sqrt[4-(2sinx)^2]
     
  8. Jul 22, 2008 #7

    Dick

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    You forgot the part of the integrand that's coming from the dx in your substitution.
     
  9. Jul 22, 2008 #8

    LHC

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    Alright, I shall try to fix it. (Gotta go, it's time for dinner.) Thanks for your help!
     
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