Question

Evaluate: ${\int }_{-1}^1{e}^{x}dx$ using limit of a sum.

• A. $\frac{e^2-1}{e}$
• B. $\frac{e^2-2}{e}$
• C. $\frac{e^2-1}{2e}$
• D. $\frac{e^2+1}{e}$

Answer 1

The correct option is A $\frac{e^2-1}{e}$

$I={\int }_a^1{e}^{x}dx$

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

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

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

we know,

$h=\frac{1-(-1)}{n}=\frac{2}{n}$

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

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

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