Question

Evaluate the given definite integrals as limit of sums:
${\int }_{-1}^1{e}^{x}dx$

Answer 1

Let $I={\int }_{-1}^1{e}^{x}dx$

We know that
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)....f(a+(n-1)h\left)\right]$, where $h=\frac{b-a}{n}$

Here, $a=-1,b=1$, and $f\left(x\right)=e^x$
$\therefore h=\frac{1+1}{n}=\frac{2}{n}$

$\therefore I=(1+1)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(-1)+f(-1+\frac{2}{n})+f(-1+2\cdot \frac{2}{n})+....+f(-1+\frac{(n-1)2}{n})]$

$=2\underset{n\to \infty }{lim}\frac{1}{n}[e^{-1}+e^{(-1+\frac{2}{n})}+e^{(-1+2\cdot \frac{2}{n})}+....e^{(-1+(n-1\left)\frac{2}{n}\right)}]$

$=2\underset{n\to \infty }{lim}\frac{1}{n}\left[e^{-1}\{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+e^{(n-1)\frac{2}{n}}\}\right]$

which forms a G.P.

$=2\underset{n\to \infty }{lim}\frac{e^{-1}}{n}\left[\frac{{\left(e^{\frac{2}{n}}\right)}^n-1}{e^{\frac{2}{n}}-1}\right]$ (Sum of n terms of G.P.)

$=e^{-1}\times 2\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2-1}{e^{\frac{2}{n}}-1}\right]$

$=(e^2-1)2e^{-1}\underset{n\to \infty }{lim}\frac{1}{n\left(\frac{2}{n}\right)}\left[\frac{1}{\frac{e^{\frac{2}{n}}-1}{\frac{2}{n}}}\right]$

$=e^{-1}\left[\frac{2(e^2-1)}{2}\right]$ $[{lim}_{h\to 0}\left(\frac{e^h-1}{h}\right)=1]$

$=\frac{e^2-1}{e}$

$=e-\frac{1}{e}$