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Simple proof needed

  1. May 3, 2007 #1
    Prove that [tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}}[f(x) + f(a-x)]dx \\[/tex]
    Hence prove that [tex] \int_0^{\pi} x sin^6x dx = \pi \int_0^{\frac{\pi}{2}} sin^6x dx \\ [/tex] and evaluate this integral using the following reduction formula [tex] I_{m,n}= \frac{n-1}{m+n}I_{m,n-2} \\ [/tex]

    My effort:
    [tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}}f(x)dx + \int_{\frac{a}{2}} f(a-x)dx \\ [/tex]. Now
    [tex] \int_{\frac{a}{2}}^a f(x)dx = \int_{\frac{a}{2}}^a f(a-x)dx \\ [/tex] ... by a theorem. Therefore [tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}} f(x)dx + \int_{\frac{a}{2}}^a f(a-x)dx \\ [/tex]. I need help to change the integral's limits to 0 and a/2 on the second integral on the right, thanks very much.
     
  2. jcsd
  3. May 3, 2007 #2
    [tex]\int_{0}^{a}f(x)dx[/tex]
    change variable u=a-x then we have du=-dx
    [tex]-\int_{a}^{0}f(u)du=-\int_{0}^{a}f(a-x)dx[/tex]
    so now you have: [tex]\int_{0}^{a}f(x)dx=\int_{0}^{a}[f(x)-f(a-x)]/2dx[/tex]
    another change u=x-(a/2), which gives you:
    [tex]
    \int_{-a/2}^{a/2}f(u+a/2)-f(a/2-u)]/2du
    [/tex]
    this function in the integrand is an odd function, which means that this equals 2 times the integral but from 0 to a/2.
    if im mistaken, sorry for misleading.
     
    Last edited: May 3, 2007
  4. May 4, 2007 #3

    Gib Z

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    Homework Helper

    You are mistaken :( [tex]\int^a_0 f(x) dx = \int^a_0 f(a-x) dx[/tex]

    On your 2nd line of latex, Reversing the bounds as you did makes the integrals value negative, canceling the other negative.

    Jean - To finish it off, u=x - (a/2) for the last integral.
     
  5. May 4, 2007 #4
    well it should be [tex]-\int_{a}^{0}f(a-u)du=\int_{0}^{a}f(a-u)du[/tex]
    so yes, you are right, thanks for the correction.
     
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