Yesterday, I sat for my Real Analysis II paper. There I found a question asking to integrate $\displaystyle\int_0^1 xe^x \, dx$ without using antiderivatives and integrating by parts.

I tried it by choosing a partition


but I was not able to show that $\displaystyle \lim_{n \to \infty} U(f,P_n)=\lim_{n \to \infty} L(f,P_n)=1$

  • 1
    $\begingroup$ One can fairly easily, though not necessarily under exam conditions, find explicit expressions for the upper and lower sum. $\endgroup$ – André Nicolas Dec 1 '15 at 7:29
  • $\begingroup$ Then use L'Hopital $\endgroup$ – Empy2 Dec 1 '15 at 7:56
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    $\begingroup$ @AndréNicolas: in what way is your comment useful ? $\endgroup$ – Yves Daoust Dec 1 '15 at 8:28
  • $\begingroup$ Ir possibly wasn'r. I was too indirectly suggesting OP calculate the sum. $\endgroup$ – André Nicolas Dec 1 '15 at 14:05
  • $\begingroup$ Another possibility: write the power series for $e^x$, integrate $xe^x$ term-by-term, and recognize the resulting sum. $\endgroup$ – GEdgar Jan 5 '18 at 10:25

We have

$$U(f,P_n) = \frac1{n^2} \sum_{k=1}^n k e^{k/n}= L(f,P_n) + \frac{e}{n}.$$

If the limit of the upper sum exists, then it is identical to the limit of the lower sum.

Note that

$$\sum_{k=1}^nk r^k = \frac{r-r^{n+1}}{(1-r)^2}- \frac{nr^{n+1}}{1-r}.$$

Using $r = e^{1/n}$ we have as $n \to \infty$

$$U(f,P_n)= \frac{1/n}{1-e^{1/n}}\frac{1/n}{1-e^{1/n}}e^{1/n}(1-e)- \frac{1/n}{1-e^{1/n}}e^{1/n}e \to 1,$$


$$\lim_{n \to \infty} \frac{1/n}{1-e^{1/n}}= -1$$


My answer:

$U(f,P_n)=\frac{1}{n^2}\left(e^{\frac{1}{n}}+2e^{\frac{2}{n}}+3e^{\frac{3}{n}}+...+ne^1\right)=\frac{1}{n^2}\displaystyle\sum_{k=1}^{n}ke^{\frac{k}{n}}=\frac{1}{n^2}\displaystyle\sum_{k=1}^{n}k\left(1+\frac{k}{n}+\frac{k^2}{2!n^2}+\frac{k^3}{3!n^3}+..\right)=\frac{1}{n^2} \displaystyle\sum_{k=1}^{n}k+\frac{1}{n^2}\displaystyle\sum_{k=1}^{n}k\left(\frac{k}{n}+\frac{k^2}{2!n^2}+\frac{k^3}{3!n^3}+..\right)=\frac{n(n+1)}{2n^2}+\frac{n(2n+1)(n+1)}{2!6n^3}+\frac{n^2(n+1)^2}{3!4n^4}+....$

Thus when $n\to \infty $ , $U(f,P_n)=\frac{1}{2}+\frac{1}{2!6}+\frac{1}{3!4}+\frac{1}{4!30}+..$ should be 1


My attempt was wrong. Fixing it with my choice of partition unnecessarily complicates the calculation, especially when a solution exists with the partition the OP selected.

SUGGESTION: What would happen if you change the partition to $P_n=[0,\frac{1}{n},\frac{1}{n-1},\ldots,\frac{1}{3},\frac{1}{2},1]$?

  • $\begingroup$ Right. I was simply summing $f(x)$ at the partition points. I overlooked the addition of $\Delta\,x$ for the integral. This unnecessarily complicates my solution especially when you found the way out with the OP's partition. $\endgroup$ – Laars Helenius Dec 1 '15 at 8:25
  • $\begingroup$ Each of the intervals in the partition needs to be able to reach arbitrary small sizes. You cannot do it with $[0, \frac{1}{n}\dots\frac{1}{2},1]$. $\endgroup$ – Element118 Dec 1 '15 at 10:08
  • $\begingroup$ Right. I had been studying absolute continuity recently and used partition tricks like my suggestion a lot. I has been a while since I have calculated a Riemann sum. My rust was showing big time! $\endgroup$ – Laars Helenius Dec 1 '15 at 10:46

Here is a rather different way of doing it. Let's assume we know the following:

$$\frac{e^t-1}t=\int_0^1e^{xt}\ dx$$

Differentiate both sides (under the integral on the right, which may be proven easily)

$$\frac{(t-1)e^t+1}t=\int_0^1xe^{xt}\ dx$$

And with $t=1$,

$$1=\int_0^1xe^x\ dx$$


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