Riemann sum calculations

Question

After writing an integral as a limit of a Riemann sum, how do we actually calculate the integral? It seems that generally, we're in some form that isn't simplified. For example, take

$$\int_0^3e^xdx=e^x|_0^3=e^3-1.$$

But this is also $$\int_0^3e^xdx=\lim_{n\to\infty}\sum_{i=1}^n\frac{3e^{3i/n}}{n}.$$

After getting to that last expression, is there anything we can do with it, or is this just some sort of way to define it? I mean, we can do $3\lim_{n\to\infty}\sum_{i=1}^n\frac{e^{3i}}{ne^n},$ but it doesn't seem like that really helps. How do we get from the last expression to $e^x+C$, or is that not even the purpose?

It would make more sense if integration and FTC calculate Riemann sums rather than the other way around, but sometimes the way it's presented is that we write integrals as Riemann sums and not the other way around.

Answer 1

It is unlikely that you can prove in a direct and simple way that$$\lim_{n\to\infty}\sum_{i=1}^n\frac{3e^{3i/n}}{n}=e^3-1.$$However, before we are able to apply the Fundamental Theorem of Calculus to compute integrals, we must know what $\int_a^bf(t)\,\mathrm dt$ means. And the introduction of Riemann sums appears naturally in the context of the definition of the Riemann integral.

Answer 2

Take a uniform partition of $[0,3]$ and consider the upper sum

$$\frac{3}{n} \sum_{k=1}^n e^{3k/n} = \frac{3}{n}\frac{e^{3/n} - e^{3(n+1)/n}}{1 - e^{3/n}} = e^{3/n} \frac{3}{n}\frac{1 - e^{3}}{1 - e^{3/n}}\\ = e^{3/n}\frac{3}{n(e^{3/n}- 1)}(e^3 - 1)$$

Since $e^{3/n} = 1 + \frac{3}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)$, we have

$$\lim_{n \to \infty}e^{3/n}= 1, \quad \lim_{n \to \infty}n(e^{3/n}-1)= 3,$$

and it follows that

$$\int_0^3 e^x \, dx = \lim_{n \to \infty}\frac{3}{n} \sum_{k=1}^n e^{3k/n} = e^3 - 1$$