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Thread: Differentiating an integral

  1. #1
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    Differentiating an integral

    Hi,

    I am unsure how to differentiate an integral. The specific problem I am working on at the moment is:

    $\displaystyle \frac{\partial }{\partial T} \int_t^T F(t;s) ds $

    s is a dummy variable used to represent T. The solution to this is

    $\displaystyle F(t;T) $

    ...but I am unsure why we would get this.

    Thanks in advance.

    Peter
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  2. #2
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    That's the first fundamental theorem of calculus: Fundamental theorem of calculus - Wikipedia, the free encyclopedia

    If there's anything you should ever remember from calculus, it's this theorem.
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  3. #3
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    Hi !
    Quote Originally Posted by peterpan View Post
    Hi,

    I am unsure how to differentiate an integral. The specific problem I am working on at the moment is:

    $\displaystyle \frac{\partial }{\partial T}\int_t^T F(t;s) ds $

    s is a dummy variable used to represent T. The solution to this is

    $\displaystyle F(t;T) $

    ...but I am unsure why we would get this.

    Thanks in advance.

    Peter
    Let $\displaystyle \tilde{F}(s)=F(t;s)$. Let $\displaystyle \mathcal{F}$ be an antiderivative of $\displaystyle \tilde{F}$.
    By definition of the integral (I think you call it "fundamental theorem of calculus"), we have :

    $\displaystyle \int_t^T \tilde{F}(s) ~ds=\mathcal{F}(T)-\mathcal{F}(t)$.

    Now, $\displaystyle \frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\frac{\partial}{\partial T} ~ \left(\mathcal{F}(T)-\mathcal{F}(t)\right)$

    Assuming that t is not a function of T, we can thus consider that $\displaystyle \mathcal{F}(t)$ is a constant with respect to T. Hence $\displaystyle \frac{\partial}{\partial T} ~\mathcal{F}(t)=0$


    $\displaystyle \therefore \frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\frac{\partial}{\partial T} ~\mathcal{F}(T)$

    But the derivative of $\displaystyle \mathcal{F}$ is $\displaystyle \tilde{F}$. We can then say that $\displaystyle \frac{\partial}{\partial T} \int_t^T F(t;s) ~ds=\frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\tilde{F}(T)=F(t;T)$
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  4. #4
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    Here's the general expression:

    $\displaystyle \frac{\partial}{\partial x}\int_{f_1(x)}^{f_2(x)}G(x,t)dt=
    G(x,f_2(x))\frac{d f_2}{dx}-G(x,f_1(x))\frac{d f_1}{dx}+\int_{f_1(x)}^{f_2(x)}
    \frac{\partial G}{\partial x}dt$

    Then:

    $\displaystyle \frac{\partial}{\partial x}\int_{4x+x^2}^{\sin(x)} (2x^2\sin(t)+t^2e^{x})dt$

    is a piece of cake right?
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