Question

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

Answer 1

We know that ${\int }_{-1}^{1}f\left(x\right)dx={lim}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+.....+f\{a+(n-1\left)h\right\}]$
where, nh =b -a
Given, ${\int }_{-1}^1{e}^{x}dx$
Here, a=-1, b=1 and nh =b -a =2
and $f\left(x\right)=e^x,thenf\left(a\right)=f\left(1\right)=e^{-1},f(-1)f(-1+h)=e^{(-1+h)}...f\left[\right(-1+(n-1)h\left)\right]=e^{(-1+(n-1\left)h\right)}$
$\therefore {\int }_{-1}^1{e}^{x}dx={lim}_{h\to 0}h[e^{-1}+e^{(-1+h)}+e^{-1+2h}+....+e^{(-1+(n-1\left)h\right)}]={lim}_{h\to 0}h{e}^{-1}[1+e^h+e^{2h}+....+e^{(n-1)h}][\because a+ar+a{r}^2+....+a{r}^n=\frac{a(r^n-1)}{(r-1)}]={lim}_{h\to 0}\frac{h}{3}\frac{\left\{\right(e^h{)}^n-1\}}{e^h-1}=\frac{1}{e}{lim}_{h\to 0}\frac{e^{hn}-1}{\frac{e^h-1}{h}}(\because |e^h|>1)$
$=\frac{1}{e}{lim}_{h\to 0}\frac{e^2-1}{\frac{e^h-1}{h}}=\frac{1}{e}\left(\frac{e^2-1}{1}\right)(\because {lim}_{h\to 0}\frac{e^x-1}{x}=1andnh=2)=\frac{e^2-1}{e}=e-\frac{1}{e}$