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Math Help - Differentiating an Integral

  1. #1
    Newbie emterics90's Avatar
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    Differentiating an Integral

    Hi, everyone, I'm new here and don't know how to type mathematics, but I have a scanner.



    I have a function L_A and it is an integral. I want to differentiate this function with respect to A. I already have the answer written but what I don't know is how it was obtained.

    Just by looking at the answer I can sort of see some sort of pattern, and I have written what I think is some sort of rule on the second half of this page, but I still don't really know what kind of differentiation rule is used here, so if any smart people here know it would greatly help me thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emterics90 View Post
    Hi, everyone, I'm new here and don't know how to type mathematics, but I have a scanner.



    I have a function L_A and it is an integral. I want to differentiate this function with respect to A. I already have the answer written but what I don't know is how it was obtained.

    Just by looking at the answer I can sort of see some sort of pattern, and I have written what I think is some sort of rule on the second half of this page, but I still don't really know what kind of differentiation rule is used here, so if any smart people here know it would greatly help me thanks!
    All the variables you have are confusing me. You want to integrate with respect to A, yet you have L_A as a function of s, anyway, I believe you want to use The Second fundamental theorem of calculus.

    Second Fundamental Theorem of Calculus:

    If we have a function defined as F(x) = \int_{c}^{u} f(t) ~dt

    where f(t) is a continuous function, c is a constant, and u is a function of x

    Then the function f(x) = F'(x) (with respect to x) is given by:

    f(x) = \frac {dF}{dx} = \frac {d}{dx} \int_{c}^{u} f(t)~dt = f(u) \cdot u'

    In short, f(x) = f(u) \cdot u'



    Now on to your question. I will use B and C to be the same functions of A as you have:

    L_A = \int_{C}^{B} e^{-r(t - A)} W_t ~dt

    \Rightarrow L_A = \int_{C}^{0} e^{-r(t - A)} W_t ~dt + \int_{0}^{B} e^{-r(t - A)} W_t ~dt

    \Rightarrow L_A = - \int_{0}^{C} e^{-r(t - A)} W_t ~dt + \int_{0}^{B} e^{-r(t - A)} W_t ~dt

    Now we can apply the second fundamental theorem to each of those integrals since they are in the required form.

    \Rightarrow \frac {d L_A}{dA} = \frac {d}{dA} \int_{0}^{B} e^{-r(t - A)} W_t ~dt - \frac {d}{dA} \int_{0}^{C} e^{-r(t - A)} W_t ~dt

    \Rightarrow \frac {d L_A}{dA} = \left( e^{-r(B - A)} W_B \right) \cdot B' - \left( e^{-r(C - A)} W_C \right) \cdot C'

    \Rightarrow \frac {d L_A}{dA} = B' e^{-r(B - A)} W_B - C' e^{-r(C - A)} W_C

    And now you can plug in B, B', C and C' in their right places. Is that the answer you're looking for?
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  3. #3
    Newbie emterics90's Avatar
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    I've read up on the Second Fundamental Theorem of Calculus and had a go at just using it and here's what I got so far:




    Unfortunately, the algebra just got really big and scary, and I have a feeling I've made a mistake, so I'll get back to it later. I think I should try to do what you did and use Bs and Cs.
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  4. #4
    Newbie emterics90's Avatar
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    You got \frac {d L_A}{dA} = B' e^{-r(B - A)} W_B - C' e^{-r(C - A)} W_C<br />

    The back-of-the-book solution, when using B and C, had an extra term \int_{C}^{B}re^{r(t-A)}W_{t}dt. I think I understand your reasoning, so maybe the back-of-the-book solution is wrong or maybe when I sub the terms in that extra term disappears.
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