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How would I solve this integral?

  • Thread starter armolinasf
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  • #1
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Homework Statement



suppose [tex]\int^{1}_{0}[/tex] f(x)dx=k. Evaluate [tex]\int^{1}_{0}[/tex] xf(1-x[tex]^{2}[/tex). Give your answer in terms of k.




The Attempt at a Solution



I'm not too sure how to go about solving this one. I think there's some substitution involved but like I said I'm not sure. Thanks for the help
 

Answers and Replies

  • #2
Dick
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u=1-x^2 looks like a good bet. Try it.
 
  • #3
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That would give me
[tex]\int^{1}_{0}[/tex][tex]\sqrt{1-u}[/tex]f(u)

But where would I go from there?
 
  • #4
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By the way, I think your question is to calculate: ∫(from 0 to 1)xf(1-x^2)dx.
just let u=1-x^2, so xdx=d(0.5x^2)=-0.5du
so the integral is solved.
 
  • #5
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How would I express that in terms of k?
 
  • #6
SammyS
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That would give me
[tex]\int^{1}_{0}[/tex][tex]\sqrt{1-u}[/tex]f(u)

But where would I go from there?
You forgot the du.

[tex]u=1-x^2\quad\to\quad du=-2x\,dx[/tex]

So, [tex]xf(1-x^2)\,dx=-(1/2)f(1-x^2)\,(-2x)\,dx=f(u)\,du[/tex]
 
  • #7
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How about this: since Int (from 0 to 1) f(x)dx=k and Int (from 0 to 1) xf(1-x^2) when we let u=1-x^2 becomes Int (from 0 to 1) sqrt(1-u)f(u), wouldn't that be saying sqrt(1-u)*Int (from 0 to 1) f(u), which is no different then Int (from 0 to 1) f(x)dx=k. So the answer would be sqrt(1-u)*k, or sqrt(1-(1-x^2))*k= sqrt(x^2)*k = xk?
 
  • #8
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So then including the -1/2 it would be -xk/2

Edit: I understand this now: I got (-1/2)k as an answer
 
Last edited:
  • #9
Dick
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So then including the -1/2 it would be -xk/2

Edit: I understand this now: I got (-1/2)k as an answer
You've maybe partially got it. But that's not quite right. Don't forget to change the limits on the u integral.
 

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