Hi, i can't solve this, please if someone can solve and show me the steps (:

P is the partition.

Thanks for all the answers (:

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- August 22nd 2012, 06:57 PMSimpleManIntegrate, Riemann.
Hi, i can't solve this, please if someone can solve and show me the steps (:

P is the partition.

Thanks for all the answers (: - August 23rd 2012, 04:40 AMemakarovRe: Integrate, Riemann.
- August 29th 2012, 09:18 AMscubastevezRe: Integrate, Riemann.
so let ln(x) be f

^{-1}(x) so e^{x}=f(x)

F^{-1}(x)=xf^{-1}(x)-F(f^{-1}(x))

so http://latex.codecogs.com/png.latex?...} \ln (x) dx $= x(ln(x)- e^{ln(x)}= xln(x)-x evaluated at e and 1

which should be 1

if you haven't seen the equation before look at the geometry between a function and its inverse... they add up to make a rectangle.

I also don't know what the partition thingamabobs are but the answer is 1 - August 29th 2012, 10:11 AMHallsofIvyRe: Integrate, Riemann.
Yes, that is easy to integrate but I suspect Simple Man is asking about the Riemann sum that will give that integral. The partition is . Those values for x make it easy to find the corresponding "f(x)": . However, it is also makes the interval lengths variable- the length of the first interval is , the second , etc.

The Riemann sum will be which is, of course, the same as