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Leibniz' Integral rule

  1. Dec 1, 2007 #1
    [SOLVED] Leibniz' Integral rule

    1. The problem statement, all variables and given/known data

    Use the Leibniz' integral rule for differentiating under the integral sign to determine constants a and b such that the integral [tex]\int^{1}_{0}(ax+b-x^{2})^{2} dx [/tex] is as small as possible.

    2. Relevant equations

    Leibniz' Interation was found at http://mathworld.wolfram.com/LeibnizIntegralRule.html and is as follows:

    [tex]\frac{\partial}{\partial z} \int^{b(z)}_{a(z)} f(x,z) dx = \int^{b(z)}_{a(z)} \frac{\partial f}{\partial z} dx + f(b(z),z) \frac{\partial b}{\partial z} - f(a(z),z) \frac{\partial a}{\partial z} [/tex]

    3. The attempt at a solution

    On my integral the limits of integration are constants and therefore the integral breaks down to:

    [tex]\frac{\partial}{\partial z} \int^{b(z)}_{a(z)} f(x,z) dx = \int^{b(z)}_{a(z)} \frac{\partial f}{\partial z} dx [/tex]

    I took the partial with respect to x and I got:

    [tex]f_{x} = 2 (ax +b - x^{2})(2x-a)[/tex]=0

    So two Equations result from that: [tex]ax+b-x^{2}=0[/tex] AND [tex]2x-a=0[/tex]

    Which gives: [tex]x=\frac{a}{2}[/tex]

    I substituted that into the other equation and got


    I know this seems like some algebra somputations but I am really asking to see if I did the Leibniz integration underneath the integral correctly since I am stuck. Thanks for the help in advance.
  2. jcsd
  3. Dec 1, 2007 #2


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    Homework Helper

    Did you remember to integrate fx from 0 to 1 before setting the whole thing to zero?
  4. Dec 1, 2007 #3
    Duh, no I didn't integrate. Lol, I feel kinda dumb after that one. Thanks
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