How to calculate $\int_0^1e^xdx$ using Riemann sums.

Question

I need to calculate $\int_0^1 e^x dx$ using Riemann sums. And I'm getting stuck at the point where my intervals are $1/n$ and my final limit is $\lim_{n \to \infty} \frac1n (1/(1-e^{1/n}))$ and at this point I don't really know what to do next.

Answer 1

As I interpret your result, you have "recognized" in the Riemann sum :

$$\frac1n\left(e^{\tfrac0n}+e^{\tfrac1n}+e^{\tfrac2n}+\cdots+e^{\tfrac{(n-1)}{n}}\right)$$

the sum of a geometric progression ; indeed, setting $a=e^{\tfrac1n}$, we have

$$S_n=\frac1n(a^0+a^1+a^2\cdots+a^{n-1})$$

But at this point, you have considered that you have an infinite summation, whereas in fact, we have a finite sum with formula :

$$S_n=\frac1n \frac{a^n-1}{a-1}=\frac{\frac1n}{e^{\tfrac1n}-1}(e-1)$$

The limit of $S_n$ when $n \to \infty$ is $e-1$ (as awaited), using a classical limit. Do you see which one ?