Given that $\int_{a}^{b} e^x dx$ exists, evaluate it by using the formula $$ 1 + r + r^2 + ... + r^n = \frac{1-r^{n+1}}{1-r}, r \neq 1$$ to calculate certain Riemann sums.

As $\int_{a}^{b} e^x dx$ exists denoted $L$, we have a Riemann sum $\sigma$ over a partition $P$ of $[a,b]$ such that $\forall \epsilon>0$ , there is a $\delta>0$ where: $$|\sigma - L|<\epsilon , \,\, ||P||<\delta$$

Given the partition $P=\{a=x_0, x_1, ..., x_n = b\}$ of $[a,b]$ with some points $d_1, ..., d_n$ such that $$x_{j-1} \leq d_j \leq x_j$$ We have the Riemann sum $$\sigma= \sum_{j=1}^{n} f(d_j) (x_j-x_{j-1}) = \sum_{j=1}^{n} e^{d_j} (x_j-x_{j-1})$$

As the goal is to calculate "certain" Riemann sums, let's evaluate the lower sum of the partition: As $d$ will be fixed on the left hand side of each subinterval so that $d_j-1=x_{j-1} = \{x_0, x_1, ..., x_{n-1}\}$ or $ \{0,1,2,...,n-1\}$ We have $$\sum_{j=1}^{n-1} f(d_{j-1})= e^0 + e^1 ... +e^{n-1} = \frac{1-e^{(n-1)+1}}{1-e^1}$$

the lower sum:

$$s(P)= \sum_{j=1}^{n} e^{d_{j-1}} (x_j-x_{j-1})$$ $$s(P) = (b-a) \frac{1-e^{(n-1)+1}}{1-e^1} $$ $$s(P) = (b-a) \frac{1-e^{n}}{1-e} $$

Evaluating the upper sum, $d_j= \{1,2,3, ..., n\}$

$$S(P)= \sum_{j=1}^{n} e^{d_j} (x_j-x_{j-1}) =(b-a) \left[ \frac{1-e^{n+1}}{1-e} -1 \right]= (b-a) \frac{e(1-e^n)}{1-e}$$

With $m= \inf \{f(d_j):x \in [a,b]\}$ and $M= \sup \{f(d_j):x \in [a,b]\}$,the goal is to bound lower and upper Riennam integrals such as $$m(b-a) \leq s(P) \leq \underline{\int}_a^b \leq \int_a^b \leq \overline{\int}_a^b \leq S(P) \leq M(b-a)$$ $$e^a(b-a) \leq (b-a) \frac{1-e^{n}}{1-e} \leq \underline{\int}_a^b \leq \int_a^b \leq \overline{\int}_a^b \leq (b-a) \frac{e(1-e^n)}{1-e} \leq e^b(b-a)$$

This is so far what I understood from the evaluation of the Riemann integral using lower and upper integrals. I did not see yet the subtopic on the limit with the integral. I am not so sure if I am going in the right direction with what I did above. What needs to be done to complete this work? Thx in advance for your feedback or explanation.


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

$$\begin{align}\frac{b-a}{n} \sum_{k=1}^n e^{a + (b-a)k/n} &= e^a\frac{b-a}{n} \sum_{k=1}^n e^{(b-a)k/n} \\ &= e^a\frac{b-a}{n}\frac{e^{(b-a)/n} - e^{(b-a)(n+1)/n}}{1 - e^{(b-a)/n}} \\ &= e^a e^{(b-a)/n} \frac{b-a}{n}\frac{1 - e^{b-a}}{1 - e^{(b-a)/n}}\\ &= e^{(b-a)/n}\frac{b-a}{n(e^{(b-a)/n}- 1)}(e^b - e^a)\end{align}$$

The integral (since it is assumed to exist) equals the limit of any Riemann sum as $\|P \| \to 0$ (and $n \to \infty$ for the uniform partition.)


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

we have

$$\int_a^b e^x \, dx = \lim_{n \to \infty}\frac{b-a}{n} \sum_{k=1}^n e^{a + (b-a)k/n} = e^b - e^a$$


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