Integrate, Riemann.

**SimpleMan** · August 22nd 2012, 05:57 PM

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 (:

**emakarov** · August 23rd 2012, 03:40 AM

Originally Posted by SimpleMan

Hi, i can't solve this, please if someone can solve

What do you mean by "solve" here?

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 $[q^{0};q^{1};q^{2};...;q^{n}]$ ; $\displaystyle q=e^{\frac{1}{n}}$ ?

**scubastevez** · August 29th 2012, 08:18 AM

so let ln(x) be f^-1(x) so e^x=f(x)
F^-1(x)=xf^-1(x)-F(f^-1(x))
so $\displaystyle \int_{1}^{e} \ln (x) dx$ = 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

**HallsofIvy** · August 29th 2012, 09:11 AM

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
$\frac{1}{n}(ne- e^{(n-1)/n}- \cdot\cdot\cdot- e^{2/n}+ e^{1/n})$

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