Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# Evaluate the definite integral as limits of sums$\int\limits_{-1}^1e^xdx$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• $\int \limits_a^b f(x)dx=\lim_ {h \to 0} h[f(a)+f(a+h)+.....f(a+(n-1)h]$
• Where $h=\large\frac{b-a}{n}$
• (ii)$\lim _{h \to 0} \large\frac{e^h-1}{h}=1$
• (iii) $a+ar+ar^2+.....ar^{n-1}=a\bigg(\large\frac{r^n-1}{r-1}\bigg)$
Given $\int \limits_ {-1}^1 e^x dx$

$\int \limits_a^b f(x)dx=\lim_ {h \to 0} h[f(a)+f(a+h)+.....f(a+(n-1)h]$

$f(x)=e^x ;\; b=1\;a=-1$

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

Hence $\int \limits_ {-1}^1 e^x dx$

$=\lim_ {h \to 0} h[f(-1)+f(-1+h)+.....f(-1+(n-1)h]$

$=\lim_ {h \to 0} h[e^{-1}+e^{-1+h}+e^{-1+2h}.....e{-1+(n-1}h]$

$=\lim_ {h \to 0} h[e^{-1}+e^{-1+h}.e^h+e^{-1}.e^{2h}.....e{-1}.e^{(n-1)h}]$

Taking $e^{-1}$ as common factor

$=\lim_ {h \to 0} h.e^{-1}[1+e^h+e^{2h}+.....e{(n-1}h]$

This is a geometric progression so for the sum of the series we can apply a $a \bigg(\large\frac{r^n-1}{r-1}\bigg)$

Here a=1,r=h.

Therefore $1+e^h+e^{2h}+.....e{(n-1}h=\large\frac{(e^h)^n-1}{e^h-1}$

Therefore $\int \limits_ {-1}^1 e^x dx =\lim_ {h \to 0}he^{-1}\bigg[\large\frac{(e^h)^n-1}{e^h-1}\bigg]$

But $\lim_{h \to 0} \large\frac{e^h-1}{h}=1$

Now substituting for $h=\large\frac{2}{n}$

nh=2

Hence $\int \limits_{-1} ^1 e^xdx=e^{-1}\bigg[\large\frac{e^2-1}{\frac{e^h-1}{h}}\bigg]$

$=e^{-1}(e^2-1)$

$=e-e^{-1}$

answered Feb 9, 2013 by

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer