Thread: Derivative of an integral

1. Derivative of an integral

Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x

2. Originally Posted by AlbatrossIV
Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
How is u related to x?

3. thats where my confusion comes from, I'm not sure how to incorporate the u's and x's

4. See attacment

5. Originally Posted by AlbatrossIV
Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
So the problem is
$\frac{d}{dx}\int_{9x}^{3x} \frac{u+5}{u-8} du$?
You should know that, by the "fundamental theorem of calculus", $\frac{d}{dx}\int_a^x f(t)dt= f(x)$
It is a little bit more complicated by the fact that the limits of integration are 9x and 3x but you can do this:
$\int_{9x}^{3x} \frac{u+5}{u-8}du= \int_0^{3x}\frac{u+5}{u-8}du- \int_0^{9x} \frac{u+5}{u-8}du$

And the fact that the upper limit is not just "x" can be handled by setting y= 3x in the first integral and y= 9x in the second:
The derivative of $\int_0^y \frac{u+5}{u-8}du$ [b]with respect to y[/tex] is just $\frac{y+5}{y-8}$ and the derivative with respect to x is (by the chain rule) that times the derivative of y with respect to x.