How to Differentiate an Integral with a Variable Upper Limit?

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SUMMARY

The discussion focuses on differentiating the integral equation \(\int_t^T n(s) (1- e^{-c (T-s)}) ds = c F(T)\) with respect to \(T\). The result obtained is \(\int_t^T n(s) ( e^{-c (T-s)}) ds = \frac{\partial F(T)}{\partial T}\). The fundamental theorem of calculus is applied, specifically the rule that states \(\frac{d}{dT}\int_{a}^{T}g(s)ds=g(T)\), to simplify the integral and clarify the dependence on \(T\).

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dbb04
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I have this equation


[tex] \int_t^T n(s) (1- e^{-c (T-s)}) ds = c F(T)[/tex]


and I need to differentiate both sides with respect to T

[tex] \frac{\partial }{\partial T}[/tex]

to get the following result

[tex] \int_t^T n(s) ( e^{-c (T-s)}) ds = \frac{\partial F(T)}{\partial T}[/tex]

How was it done ? What integration and differentiation rule was used ? If you could show it step by step I would appreciate.
 
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I would rewrite the integral:

[tex] \int_t^T n(s) (1- e^{-c (T-s)}) ds = \int_t^T n(s)ds-e^{-cT}\int_t^T n(s) e^{cs}} ds[/tex]

Then use the tried and true fundamental theorem of calculus (assuming g is continuous):

[tex]\frac{d}{dT}\int_{a}^{T}g(s)ds=g(T)[/tex]

The purpose of rewriting was to remove any potentially confusing dependence of T from the integrands.
 
Yeah, sure. Now I see it.

Thanks very much for the prompt reply
 

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