Integral using the Riemann Sum

Question

Here I have a question to to calculate $$\int_{0}^{2}e^{x}dx$$ using Riemann sums. I got to the part where $$\int_{0}^{2}e^{x}dx=\lim....(e^{\frac{2i}{n}})\frac{2}{n}=\lim....(e^{\frac{2(\frac{r-r^{n+1}}{1-r})}{n}})\frac{2}{n}$$ What should I do now? Can I expand this equation further? It doesnt seem like it. Thank you for helping.

Answer 1

Yea, so we are partitioning $[0,2]$ into $n$ equal pieces each of length $\frac{2}{n}$. In particular:

$$x_i = 0+\frac{2i}{n} = \frac{2i}{n}$$

Then, we pick our evaluation set $T = \{x_i: i \in \{1,2,\ldots,n\}\}$. So, consider the Riemann Sum:

$$R(f,P,T) = \sum_{i=1}^{n} f(x_i) \Delta x_i = \sum_{i=1}^{n} e^{\frac{2i}{n}} \frac{2}{n}$$

Observe that $\sum_{i=1}^{n} \left(e^{\frac{2}{n}} \right)^i$ is a geometric sum with common ratio $e^{\frac{2}{n}}$. So, we have:

$$\sum_{i=1}^{n} \left(e^{\frac{2}{n}} \right)^i = \frac{1-\left(e^{\frac{2}{n}}\right)^{n+1}}{1-e^{\frac{2}{n}}}$$

Observe that the numerator of this expression simplifies to $1-e^{2+\frac{2}{n}}$. It is easy to see that this expression goes to $1-e^2$ as $n \to \infty$. Then, study the expression:

$$\frac{\frac{2}{n}}{1-e^{\frac{2}{n}}}$$

It is relatively easy to see that as $n \to \infty$, this expression goes to $-1$ (try proving this). So, it is the case that:

$$\lim_{n \to \infty} R(f,P,T) = e^2-1$$

and we are done. Now, alternatively, if you don't want to go that route with it, then there's a faster way to do the problem without having to do this sort of gymnastics.

Since $x \mapsto e^x$ is continuous on $[0,2]$, it is Riemann Integrable there. Let $P = \{0 = x_0 < x_1 < \ldots < x_{n-1} < x_n = 2\}$ be any partition of $[0,2]$. Since $x \mapsto e^x$ is continuous and differentiable on every interval of $\mathbb{R}$:

$$\forall i: \exists t_i \in (x_{i-1},x_i): e^{t_i} = \frac{e^{x_i}-e^{x_{i-1}}}{\Delta x_i}$$

This is just a consequence of the MVT. Then, pick $T = \{t_i: i \in \{1,2,\ldots,n\}\}$ as your evaluation set and compute the Riemann Sum:

$$R(f,P,T) = \sum_{i=1}^{n} f(t_i) \Delta x_i = \sum_{i=1}^{n} \left(e^{x_i}-e^{x_{i-1}} \right) = e^2-e^0 = e^2-1$$

where we have made use of the fact that we have a telescoping sum. This approach is nice because it actually generalizes and gives us a pretty neat proof of the Fundamental Theorem of Calculus :D

Answer 2

\begin{align} \int_0^2e^xdx&=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac2{n}e^{2i/n}\\ &=\lim_{n\to\infty}\frac2{n}\left(\frac{e^{2}-1}{e^{2/n}-1}\right)\\ &=e^2-1. \end{align}