Question

Evaluate $\stackrel{2}{\underset{0}{\int }}{e}^{x}dx$ as a limit of sum.

Answer 1

Consider, $I={\int }_0^2{e}^{x}dx$

${\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{lim}h\left(f\right(a)+f(a+h)+....f(a+(n-1)h)$

$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left(f\right(a)+f(a+h)+....f(a+(n-1)h)$

$h=\frac{(b-a)}{n}$

${\int }_0^2{e}^{x}dx=(2-0)\underset{n\to \infty }{lim}\frac{1}{n}(e^0+e^{\frac{2}{n}}+e^{\frac{8}{n}}+.....e^{2n-\frac{2}{n}})$

Using sum to n terms of a GP

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

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