Prove $∫e^x \, dx=e-1$ using rectangular method

Question

I need to prove the following using rectangular method:

$$∫_0^1e^x\,dx=e-1.$$

Otherwise speaking, I need to solve integral via area of ​​curvilinear trapezoid. And I have no idea how to do that. Would really appreciate your help.

Answer 1

I'm guessing you mean this is to be done by limits of Riemann sums.

If $x$ goes from $0$ to $1$ by steps of $\Delta x=1/n,$ then at the $i$th step we have $x=0 + i/n,$ and so

$$ \int_0^1 e^x\,dx = \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n e^{0 \,+\, i/n}. $$ And \begin{align} \frac 1 n \sum_{i=1}^n e^{0\,+\,i/n} = {} & \frac 1 n \times \Big( \text{a sum of a geometric series} \Big) \\[10pt] = {} & \frac 1 n \times\text{first term} \times \left( \frac{(\text{common ratio})^{\text{number of terms}}-1}{(\text{common ratio})-1} \right) \\[10pt] = {} & \frac 1 n \cdot e^{1/n} \left( \frac{e^{n/n}-1}{e^{1/n}-1} \right) \\[10pt] = {} & (e-1) \cdot e^{1/n} \cdot \frac 1 {n(e^{1/n}-1)} \end{align} And then we have $e^{1/n}\to1$ as $n\to\infty,$ and $\dfrac 1 {n(e^{1/n} - 1)} \to 1 $ as $n\to\infty.$

There are a number of ways to establish that last limit. Here's one: $$ \lim_{n\to\infty} \frac{e^{1/n}-1}{1/n} = \lim_{\Delta x\,\to\,0} \frac{e^{0\,+\,\Delta x} - e^0}{\Delta x} = \lim_{\Delta x\,\to\,0} \frac{\Delta e^x}{\Delta x} = \left. \frac d {dx} e^x \right|_{x=0} = e^0 = 1. $$

Answer 2

You could try this : let $\delta x = \frac{1}{n}$, then $x_{k} = \frac{k}{n}$. Now we can consider $\displaystyle \sum_{k=0}^{n-1}\frac{1}{n}(e^{x_{k+1}}-e^{x_{k}})$, then consider $\displaystyle\sum \frac{e^{k/n}(e^{1/n}-1)}{n}$. Now estimate $e^{1/n}$ and find the ''$\lim\sum$''