Using definition of definite integral to prove this equation How do I use the definition of the integral to prove the following?
$$\int_a^b e^x\,dx=e^b-e^a$$
 A: $$\begin{align}
\int_a^b e^x \, dx &= \lim_{n \to \infty} \frac{b-a}{n}\sum_{k=0}^{n} e^{a+k(b-a)/n} \\
&= e^a \lim_{n \to \infty} \frac{b-a}{n}\sum_{k=0}^{n} \left(e^{(b-a)/n}\right)^k \\
&= e^a \lim_{n \to \infty} \frac{b-a}{n} \frac{e^{(n+1)(b-a)/n}-1}{e^{(b-a)/n}-1} \\
&= e^a \lim_{n \to \infty}  \left(e^{b-a}e^{(b-a)/n}-1\right)\frac{(b-a)/n}{e^{(b-a)/n}-1}
\end{align}$$
We can use the Taylor series of the exponential here to evaluate the limit:
$$ \lim_{x \to 0} \frac{x}{e^x-1} = \lim_{x \to 0} \frac{x}{1+x+O(x^2)-1} = 1, $$
and so the answer is
$$ \int_a^b e^x \, dx = e^a \lim_{n \to \infty}  \left(e^{b-a}e^{(b-a)/n}-1\right) = e^b-e^a. $$
A: Here is an approach that avoids summing a geometric series, but relies
on the fact that the integral exists.
Suppose $P = (1=x_0,x_1,..., x_n=b)$ is a partition of $[a,b]$. Note that
$f'(x) = f(x) $ and $f(x) >0$, so $f$ is increasing.
Then
$L(f,P) = \sum_k f(x_k) (x_{k+1}-x_k) \le \sum_k f(\xi_k) (x_{k+1}-x_k) \le \sum_k f(x_{k+1}) (x_{k+1}-x_k) = U(f,P)$ for any $\xi_k \in [x_{k+1},x_k]$.
From the mean value theorem, we can find $\xi_k$ such that
$f(x_{k+1})-f(x_k) = f'(\xi_k) (x_{k+1}-x_k)$, hence
$\sum_k f(\xi_k) (x_{k+1}-x_k) = f(b)-f(a)$ and so
$L(f,P) \le f(b)-f(a) \le U(f,P)$.
Taking limits as $\operatorname{mesh} P$ becomes small gives the desired
result.
