Certain step in proving integral of $e^x$ using Riemann-sums.

Question

I'm reading a script right now, in which $$\int_{a}^{b} e^x dx = e^b-e^a$$ is proven using Riemann-sums. Unfortunately, I have trouble understanding a step in the beginning.

Some background: The subdivision of an interval $[a,b] \subset \mathbb{R}$ is put as: $[a,b] = [x_0,x_1]\cup [x_1,x_2]\cup...\cup[x_{m-1},x_m]$ with $a = x_0<x_1<x_2<...<x_{m-1}<x_m = b$.

The "fineness" (I'm not sure this is the right translation, could also be "acuteness") is:

$d(\{x_i\}) =$ max$|x_{j-1}-x_j|=$ max$\triangle x_j$.

The fulcrum in every interval $[x_{j-1}-x_j]$ is called $\xi_j$, and $f(\xi_j)$ is the value of the fulcrum.

Finally, the Riemman-sum is definded as: $$\rho(f;\{x_j\},\{\xi_j\}) = \sum_{j=1}^{m} f(\xi_j)\triangle x_j = \sum_{j=1}^{m} f(\xi_j)(x_j-x_{j-1})$$.

Now, the step I can't understand is: $$\sum_{j=1}^{m} e^{\xi_j}\triangle x_j = e^b-e^a+\sum_{j=1}^{m}(e^{\xi_j}-\frac{e^{x_j}-e^{x_{j-1}}}{x_j-x_{j-1}})\triangle x_j$$,

whereas the first step I can still follow. Before, we had taken the Riemann-sum of $f(x)=x$, in which the same expanding step felt intuitive. So if anybody has a hint on how that step is legitimate or on where I could read more about that proof, I'd be very thankful.

Answer 1

It's a simple trick of telescoping sum:

Note that $x_j-x_{j-1}=\Delta x_j$, so we have $$\sum_{j=1}^m\frac{e^{x_j}-e^{x_{j-1}}}{x_j-x_{j-1}}\Delta x_j=\sum_{j=1}^m(e^{x_j}-e^{x_{j-1}})=e^b-e^a$$

Answer 2

Note that $$e^b-e^a+\sum_{j=1}^{m}(e^{\xi_j}-\frac{e^{x_j}-e^{x_{j-1}}}{x_j-x_{j-1}})\triangle x_j=e^b-e^a+\sum_{j=1}^{m}e^{\xi_j}\triangle x_j -\sum_{j=1}^m\frac{e^{x_j}-e^{x_{j-1}}}{x_j-x_{j-1}}\triangle x_j=e^b-e^a+\sum_{j=1}^{m}e^{\xi_j}\triangle x_j -\sum_{j=1}^m{e^{x_j}-e^{x_{j-1}}}=\sum_{j=1}^{m}e^{\xi_j}\triangle x_j$$ Was that that you didn’t understand?