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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral as limits of sums\[\int\limits_{-1}^1e^xdx\]

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  • $\int \limits_a^b f(x)dx=\lim_ {h \to 0} h[f(a)+f(a+h)+.....f(a+(n-1)h]$
  • 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 $h=\large\frac{2}{n}$
Hence $\int \limits_{-1} ^1 e^xdx=e^{-1}\bigg[\large\frac{e^2-1}{\frac{e^h-1}{h}}\bigg]$



answered Feb 9, 2013 by meena.p
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