# Integrate, Riemann.

• Aug 22nd 2012, 05:57 PM
SimpleMan
Integrate, Riemann.
Hi, i can't solve this, please if someone can solve and show me the steps (:
$\displaystyle \int_{1}^{e} \ln (x) dx$ P is the partition. $P:[q^{0};q^{1};q^{2};...;q^{n}] ; \displaystyle q=e^{\frac{1}{n}}$

Thanks for all the answers (:
• Aug 23rd 2012, 03:40 AM
emakarov
Re: Integrate, Riemann.
Quote:

Originally Posted by SimpleMan
Hi, i can't solve this, please if someone can solve

What do you mean by "solve" here?

Quote:

Originally Posted by SimpleMan
$\displaystyle \int_{1}^{e} \ln (x) dx$ P is the partition. $P:[q^{0};q^{1};q^{2};...;q^{n}] ; \displaystyle q=e^{\frac{1}{n}}$

What do you mean by $[q^{0};q^{1};q^{2};...;q^{n}] ; \displaystyle q=e^{\frac{1}{n}}$?
• Aug 29th 2012, 08:18 AM
scubastevez
Re: Integrate, Riemann.
so let ln(x) be f-1(x) so ex=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)- eln(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
• Aug 29th 2012, 09:11 AM
HallsofIvy
Re: 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 $e^{0/n}=1, e^{1/n}, e^{2/n}, ..., e^{n/n}= e$. Those values for x make it easy to find the corresponding "f(x)": $ln(1)= 0, ln(e^{1/n})= \frac{1}{n}, ln(e^{2/n})= \frac{2}{n}, ..., ln(e)= 1$. However, it is also makes the interval lengths variable- the length of the first interval is $e^{1/n}- 1$, the second $e^{2/n}- e^{1/n}$, etc.

The Riemann sum will be $\frac{e^{2/n}- e^{1/n}}{n}+ \frac{2(e^{3/n}- e^{2/n})}{n}+ \frac{3(e^{4/n}- e^{3/n})}{n}+ \cdot\cdot\cdot+ \frac{n(e- e^{(n-1)/n}}{n}$ which is, of course, the same as
$\frac{1}{n}(ne- e^{(n-1)/n}- \cdot\cdot\cdot- e^{2/n}+ e^{1/n})$