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Thread: Derivative of integral

  1. #1
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    Derivative of integral

    I got stuck on this problem. Hope anyone could give some help.
    Let $\displaystyle f:R \rightarrow R$ be continuous and $\displaystyle \epsilon >0$. Define $\displaystyle g(t)=\int_{t-\epsilon}^{t+\epsilon}f(x)dx$ for all $\displaystyle t \in R$. Show that g is differentiable and find g'
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by jackie View Post
    I got stuck on this problem. Hope anyone could give some help.
    Let $\displaystyle f:R \rightarrow R$ be continuous and $\displaystyle \epsilon >0$. Define $\displaystyle g(t)=\int_{t-\epsilon}^{t+\epsilon}f(x)dx$ for all $\displaystyle t \in R$. Show that g is differentiable and find g'
    Use any fixed real number $\displaystyle t_0$ to get
    $\displaystyle g(t)=\int_{t_0}^{t+\epsilon}f(x)dx+\int_{t-\epsilon}^{t_0}f(x)dx$
    Now do a linear substitution that eliminates $\displaystyle \epsilon$ from the upper and lower limit of the integral, respectively,
    $\displaystyle =\int_{t_0-\epsilon}^tf(z+\epsilon)dz+\int_{t}^{t_0+\epsilon} f(z-\epsilon)dz=\int_{t_0-\epsilon}^tf(z+\epsilon)dz-\int_{t_0+\epsilon}^t f(z-\epsilon)dz$
    Finally: apply fundamental theorem of calculus twice to find that
    $\displaystyle g'(t)=f(t+\epsilon)-f(t-\epsilon)$
    Last edited by Failure; Apr 14th 2010 at 06:08 AM.
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