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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following as limits of sums: $ \int\limits_0^2e^xdx$

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Toolbox:
  • $\int \limits_a^b f(x) dx= \lim_{ h \to 0} \;h\; \bigg[ f (a)+f(a+h)+f(a+2h)+....f(a+(n-1)h \bigg]$
  • where $h=\large\frac{b-a}{n}$
  • $S_n=a \bigg[\large\frac{r^n-1}{r-1}\bigg]$
$I=\int \limits^2_0 e^x dx$
We know $\int \limits_a^b f(x)dx=h\; \lim_{ h \to 0} \bigg[f(a)+f(a+h)+f(a+2h)+....f(a+(n-1)h \bigg]$
Where $h=\large\frac{b-a}{n}$
Here $a=0,\; b=2,\; f(x)=e^x$
Therefore $h=\large\frac{2-0}{n}=\frac{2}{n}$
Therefore $ \int \limits_0^2 e^x dx =\lim _{h \to 0} h \bigg[f(0)+f(h)+f(2h)+.........f(n-1)h \bigg]$
$=\lim _{h \to 0} h \bigg[e^0+e^h+e^2h+........e^{(n-1)h} \bigg]$
This is in geometric progression
Where $S_n=a \bigg[\large\frac{r^n-1}{r-1}\bigg]$
Here $r=e^h\; and\;a=e^0=1$
$\int \limits_0^2 e^x dx=\lim _{h \to 0} h \bigg[e^0\bigg\{\large\frac{(e^h)^n-1}{e^h-1}\bigg\}\bigg]$
$=\lim _{h \to 0} h \bigg[\Large\frac{e^{nh}-1}{e^h-1}\bigg]$
Since $h=\large\frac{2}{n} =>nh=2$
$=\lim _{h \to 0} h \bigg[\Large\frac{e^2-1}{e^n-1}\bigg]$
Multiply and divide the denominator by h
$=\lim _{h \to 0} h \bigg[\Large\frac{e^2-1}{\frac{(e^h-1)h}{h}}\bigg]=\frac{h}{h} \bigg[\Large\frac{e^2-1}{\frac{e^h-1}{h}}\bigg]$
But $\lim _{h \to 0}\Large\frac{e^h-1}{h}=1$
$=e^2-1$
Hence $\int \limits_0^2 e^xdx=e^2-1$

 

answered Apr 17, 2013 by meena.p
 
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