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Integration by substitution

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data

    If [tex]f[/tex] is continuous and [tex]\int^{9}_{0}f(x)dx = 4[/tex], find [tex]\int^{3}_{0}xf(x^{2})dx[/tex]

    2. Relevant equations

    None required

    3. The attempt at a solution

    Don't really know where to begin, but I tried:

    for [tex]\int^{3}_{0}xf(x^{2})dx[/tex]
    let:
    [tex]u = x^{2}[/tex]

    [tex]du = 2xdx[/tex]

    substitute

    [tex]\int^{3}_{0}xf(x^{2})dx = \int^{9}_{0}\frac{1}{2}f(u)du

    [/tex]

    NOW, for [tex]\int^{9}_{0}f(x)dx = 4[/tex]

    let
    [tex]u = x[/tex]

    [tex]du = dx[/tex]

    then we have

    [tex]\int^{9}_{0}f(x)dx = 4 = \int^{9}_{0}f(u)du[/tex]

    so now...

    [tex]\int^{9}_{0}\frac{1}{2}f(u)du = (1/2)(4) = 2[/tex]

    so our result, the answer to the second integral is 2.

    But i'm pretty sure i'm wrong, I didn't really know where else to go with this so that's what I tried. This question is in the "u-substitution" section of our text, so that's probably the method we use, some sort of substitution.

    Any help is appreciated, thank you in advance!
     
  2. jcsd
  3. Jun 2, 2010 #2

    vela

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    Your solution is correct. Why do you think it's wrong?

    The final substitution u=x is pointless, though. All that did was rename a dummy variable.
     
  4. Jun 2, 2010 #3
    it seems too simple...

    i know that the integral on [0,9] for f(x) is 4.

    when I use a "u" substitution on the integral of x(f(x^2)), I change the limits of integration, the integrand, and what we're integrating with respect to.

    so we then had to integrate on [0,9] for (1/2)f(u). But f(u) does not equal f(x) and thus int(f(u))du does not equal int(f(x))dx, so we can't mix these two functions, because they aren't the same.

    basically when i said:

    [tex]
    \int^{9}_{0}f(x)dx = 4 = \int^{9}_{0}f(u)du
    [/tex]

    that's not really true is it? since f(u) and f(x) are not the same function...

    That's kind of why I used a u=x at the end.
     
  5. Jun 2, 2010 #4

    Cyosis

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    They are the same function, f(u), f(x),f(a),f(bla) etc are all the same 'functions'. It doesn't matter what name you give to the variable. Besides from your substitution u=x it immediately follows that f(u)=f(x).
     
  6. Jun 2, 2010 #5
    that's what i figured (and I'm probably just having a brain fart b/c of the lack of sleep) but "u" does NOT equal "x"...that's what's really getting me!

    So since u ≠ x, f(u) ≠ f(x)

    blahhhh! why am i arguing if I got the right answer! Everything seems fine, it's just that one thing is bothering me! My latter u=x substitution can't be correct, since earlier I already defined u=x^2...how can i just change what "u" equals again?
     
  7. Jun 2, 2010 #6

    Cyosis

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    I can see where the confusion is coming from. You're right that if we set [itex]u=x^2[/itex] then [itex]f(u)=f(x^2) \neq f(x)[/itex]. To avoid this confusion it's better to not do another substitution with the same letter . In the integral however u functions as a dummy variable. Perhaps your confusion goes away if you consider the integral [tex]\int_0^3 s f(s^2) ds[/tex].
     
  8. Jun 2, 2010 #7
    Exactly correct....why are you getting confused
    take this as an example:
    Your friend John has 56 apples and he wants to divide these apples among 8 children.....so he gives each of the children 7 apples each....Now would the no of apples each child got change if the apples were given by you , me or any any one else....NO right....same is the case is with variables whether you take u,v or x,y,z......it really does not matter....until the function is same the limit only gets changed when you take for eg u=sqrtx....!!!!Wat you did wrong was to take u=x again it was like trying to give 7 apples to each child but this time with a person who has only 10 with him...
    Don't worry these things do take some time to sink in...
     
  9. Jun 2, 2010 #8
    I am definitely going to get some sleep and come back to this! Thanks all for your help, I'm glad I ended up with the right answer!

    since u ≠ x, it DOES matter if it's x or u!

    Alright alright, thanks again everyone, hopefully after some rest I will look at this and have an "a HAH!" moment!
     
  10. Jun 2, 2010 #9

    vela

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    The variables u and x in the integrals are dummy variables. Consider these summations:

    [tex]\sum_{i=1}^n i = \frac{n(n+1)}{2}[/tex]

    [tex]\sum_{j=1}^n j = \frac{n(n+1)}{2}[/tex]

    Your argument is like saying those two summations are different because [itex]i\ne j[/itex]. The name of the variable here really doesn't matter, though. Same thing with the integrals.
     
  11. Jun 3, 2010 #10
    hey guys, just wanted to let you all know that after I got some rest and relieved some stress, I realized where my confusion was.

    Thanks again for all your help.
     
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