Question

Obtain ${\int }_0^1{e}^{x}dx$ as the limit of sum.

Answer 1

${\int }_0^1{e}^{x}dx=(1-0)\underset{n\to \infty }{lim}\frac{1}{n}\left(f\right(0)+f(0+h)+f(0+2h)+................+f(0+(n-1)h\left)\right)$

$h=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}$

$f\left(0\right)=1,f\left(h\right)=e^h,f\left(2h\right)=e^{2h},..................f\left(\right(n-1\left)h\right)=e^{(n-1)h}$

$=\underset{n\to \infty }{lim}\frac{1}{n}(1+e^h+e^{2h}+...............e^{(n-1)h})$

$=\underset{n\to \infty }{lim}\frac{1}{n}\frac{e^{(n-1)h}-1}{e^h-1}$ $h=\frac{1}{n}$

$=\underset{n\to \infty }{lim}\frac{1}{n}\frac{e^{1-\frac{1}{n}}-1}{e^{\frac{1}{n}}-1}=\underset{n\to \infty }{lim}(e^{1-\frac{1}{n}}-1)\times (\underset{\frac{1}{n}\to 0}{lim}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}})=\underset{n\to \infty }{lim}(e^{\frac{1}{n}-1}-1)$

$=e^{(1-\frac{1}{\infty })}-1=e-1$