Derivative of a definite integral

**altave86** · September 20th 2009, 10:29 AM

What's equivalent to the following expression?

$\frac { \partial} {\partial p}\int _p^{p_0} f(\alpha,\beta)d\alpha$

**redsoxfan325** · September 20th 2009, 10:10 PM

Originally Posted by altave86

What's equivalent to the following expression?

$\frac { \partial} {\partial p}\int _p^{p_0} f(\alpha,\beta)d\alpha$

Assume that $\int f(\alpha,\beta)\,d\alpha = F(\alpha,\beta)$ . Thus,

$\int_p^{p_0}f(\alpha,\beta)\,d\alpha = F(\alpha,\beta)\big|_p^{p_0}=F(p_0,\beta)-F(p,\beta)$

Note that all operations you are doing are either w.r.t. $\alpha$ or $p$ , so $\beta$ and $p_0$ should be treated as constants, and therefore $F(p_0,\beta)$ is a constant. Call it $C$ .

Now we want $\frac{\partial}{\partial p}[C-F(p,\beta)]=\frac{\partial}{\partial p}[-F(p,\beta)]$ because the derivative of a constant is zero.

Remember how we defined $F$ though. We defined it so that $F~'=f$ . So

$\frac{\partial}{\partial p}[-F(p,\beta)]=\boxed{-f(p,\beta)}$

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