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Problems with a question

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data

    if f is a continuous function, find the value of the integral

    I = definte integral int [ f(x) / f(x) + f(a-x) ] dx from 0 to a. by making the substitution u = a - x and adding the resulting integral to I.

    this is one of the last questions in the thomas international edition calculus on substitution, for the life of me i cant see how to do this, a pointer on where to attack this would be very helpful, as i cant let it go!!!!


    2. Relevant equations



    3. The attempt at a solution

    ive obviously used the substitution which was in the question, but it doesnt really help, as when i change the variable, the resulting integral is just as bad, and i dont know how adding it to the I helps, i substituted f(x) = x, into the integral and got the answer a/2, but i cant repeat it for any other random functions. I believe a/2 is the right answer but how i got it was not the right method. help
     
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2

    Dick

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    It's not that bad. It's just confusing. If you do the suggested substitution, your integrand is f(a-u)/(f(u)+f(a-u))*(-du). But now you should realize that instead of integrating x from 0 to a, you are integrating u from a to 0. So reverse the limits of integration introducing a - sign. This kills the minus sign on the du. Now do the harmless substitution u=x (it's just a dummy variable) and add it to your original integral. What's the integrand? What's your conclusion?
     
  4. Nov 4, 2007 #3

    Dick

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    Oh, yeah, and welcome to the forums.
     
  5. Nov 4, 2007 #4
    thanks very much for the reply, i followed your advice and the integrand i had was 1, which in turn leads to a when integrated, im a little confused over the dummy variable, is u=x because the limits of the integrand are still [0-a] when reversed? also the answer in the book is a/2, does this mean i have to half the answer as i have added another integrand and if so how does this work as the new integrand is of a different value as the numerator is f(a-x) on the new one as opposed to f(x) on the original which i also verifed when i put f(x) = x^2 into the new integrand trying to understand how it actually works. or have i calculated wrong??
     
    Last edited: Nov 4, 2007
  6. Nov 5, 2007 #5
    forget that last reply, i figured it out this morning when i woke with a fresh head. the graphs have the same area but the new one comes from the other side, when f(x) = 0, then f(a-x)= a, when f(x)= a the f(a-x) = 0. painfully obvious now.
    this is website is a great idea, thanks again for your help
     
  7. Nov 5, 2007 #6

    Dick

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    You've got it. You're welcome.
     
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