Show $e^x$ is Riemann Integrable from definition, then evaluate $ \int_{0}^{2} e^x dx $

Question

My attempt:

A function is Riemann integrable if $\overline{S}_f(\mathcal{D}_n)- \underline{S}_f(\mathcal{D}_n)< \epsilon , \space \forall\epsilon>0.$

Let $\epsilon>0$. Then let $\mathcal{D}_n =\{0<\frac{1}{n}<\frac{2}{n}< \dots < \frac{2n-1}{n}<2 \}$ be a dissection.

$$ \begin{split} \overline{S}_f(\mathcal{D}_n) - \underline{S}_f(\mathcal{D}_n) &= \sum_{i=1}^{2n} \frac{e^{i/n}}{n} - \sum_{i=1}^{2n}\frac{e^{(i-1)/n}}{n} \\ &= \frac{1}{n} \sum_{i=1}^{2n} e^{i/n}(1-e^{-1/n}) \\ &= \frac{1-e^{-1/n}}{n} \sum_{i=1}^{2n} e^{i/n} \\ &= \frac{1-e^{-1/n}}{n} \sum_{i=1}^{2n}\left(e^{1/n}\right)^i \end{split} $$

This looks like a geometric series, $\sum_{i=1}^{k}x^i = \frac{x(1-x^k)}{1-x}$ so letting $x=e^{1/n}$ and $k=2n$ we get,

$$ \frac{1-e^{-1/n}}{n} \frac{e^{1/n}(1-e^{2n/n})}{1-e^{1/n}} = \frac{-(1-e^{1/n})}{n} \frac{1-e^{2n/n}}{1-e^{1/n}} = \frac{e^2 -1}{n} \to 0 \text{ as } n \to \infty $$

Thus $e^x$ is integrable, so to compute the integral we compute the upper sum.

$$ \begin{split} \overline{S}_f(\mathcal{D_n}) &= \sum_{i=1}^{2n} \frac{1}{n} e^{i/n} \\ &= \frac{1}{n} \sum_{i=1}^{2n} e^{i/n} \\ &= \frac{e^{1/n}}{n} \sum_{i=1}^{2n} e^i \\ &= \frac{e^{1/n}}{n} \frac{e(1-e^{2n})}{1-e} \end{split} $$

Then from here i'm lost, any help would be appreciated. Thanks!

Answer 1

It is better to take $$\mathcal{D}_n=\{0,\frac 2n,\frac 4n,...,2\}.$$

$$\int_0^2e^xdx=\lim_{n\to +\infty}S_n$$ with

$$S_n=\frac{2-0}{n}\sum_{k=0}^{n-1}e^{0+k\frac{2-0}{n}}$$

$$=\frac 2n\sum_{k=0}^{n-1}(e^{\frac 2n})^k$$

$$=\frac 2n\frac{1-e^2}{1-e^{\frac 2n}}$$ Now observe that

$$\lim_{n\to+\infty}\frac{\frac 2n}{1-e^{\frac 2n}}=-1$$

and finish.

Answer 2

HINT

I think your evaluation of the last series is wrong: $$ \begin{split} \overline{S}_f(\mathcal{D_n}) &= \sum_{i=1}^{2n} \frac{1}{n} e^{i/n} \\ &= \frac{e^{1/n}}{n} \sum_{i=0}^{2n-1} (e^{1/n})^i \\ &= \frac{e^{1/n}}{n} \frac{1-\left(e^{1/n}\right)^{2n}}{1-e^{1/n}}\\ &= \frac{e^{1/n}}{1-e^{1/n}} \times \frac{1-e^2}{n} \\ &= \left(1-e^2\right) e^{1/n} \frac{1/n}{1-e^{1/n}} \end{split} $$ and you compute the limits of $e^{1/n}$ and the fraction separately...