Question

$\underset{0}{\overset{2}{\int }}{e}^{x}dx$

Answer 1

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{\lim }h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)...............+f\left(a+\left(n-1\right)h\right)\right] \\ \mathrm{where}h=\frac{b-a}{n} \end{gathered}$$

$$\begin{gathered} \mathrm{Here}a=0,b=2,f\left(x\right)=e^x,h=\frac{2-0}{n}=\frac{2}{n} \\ \mathrm{Therefore}, \\ I={\int }_0^2{e}^{x}dx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+....................+f\left\{0+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[e^0+e^h+e^{2h}+.......+e^{\left(n-1\right)h}\right] \\ =\underset{h\to 0}{\lim }h\left[\frac{{\left(e^h\right)}^n-1}{e^h-1}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{e^2-1}{{\displaystyle \frac{e^h-1}{h}}}\right] \\ =\frac{e^2-1}{1} \\ =e^2-1 \end{gathered}$$