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[SOLVED] Leibniz' Integral rule
Use the Leibniz' integral rule for differentiating under the integral sign to determine constants a and b such that the integral \int^{1}_{0}(ax+b-x^{2})^{2} dx is as small as possible.
Leibniz' Interation was found at http://mathworld.wolfram.com/LeibnizIntegralRule.html" and is as follows:
\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}
On my integral the limits of integration are constants and therefore the integral breaks down to:
\frac{\partial}{\partial z} \int^{b(z)}_{a(z)} f(x,z) dx = \int^{b(z)}_{a(z)} \frac{\partial f}{\partial z} dx
I took the partial with respect to x and I got:
f_{x} = 2 (ax +b - x^{2})(2x-a)=0
So two Equations result from that: ax+b-x^{2}=0 AND 2x-a=0
Which gives: x=\frac{a}{2}
I substituted that into the other equation and got
b=\frac{a^{2}}{2}
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.
Homework Statement
Use the Leibniz' integral rule for differentiating under the integral sign to determine constants a and b such that the integral \int^{1}_{0}(ax+b-x^{2})^{2} dx is as small as possible.
Homework Equations
Leibniz' Interation was found at http://mathworld.wolfram.com/LeibnizIntegralRule.html" and is as follows:
\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}
The Attempt at a Solution
On my integral the limits of integration are constants and therefore the integral breaks down to:
\frac{\partial}{\partial z} \int^{b(z)}_{a(z)} f(x,z) dx = \int^{b(z)}_{a(z)} \frac{\partial f}{\partial z} dx
I took the partial with respect to x and I got:
f_{x} = 2 (ax +b - x^{2})(2x-a)=0
So two Equations result from that: ax+b-x^{2}=0 AND 2x-a=0
Which gives: x=\frac{a}{2}
I substituted that into the other equation and got
b=\frac{a^{2}}{2}
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.
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