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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate $ \large \int\limits_0^1e^{2-3x}\;$$dx\;$ as a limit of a sum

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  • $\int\limits_a^b f(x)dx=\lim_{h\to 0}h\{f(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h\}$
  • Where $nh=b-a$
Step 1:
Let $f(x)=e^{2-3x}$
$\int _0^1 e^{2-3x}dx=\int _0^1f(x) dx$
We know that $\int\limits_a^b f(x)dx=\lim_{h\to 0}h[\{f(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h\}]$
Where $h=\large\frac{b-a}{n}$
Here $a=0,b=1$ and $nh=1$ and $f(x)=e^{2-3x}$
$f(0)=e^2,f(0+h)=e^{2-3h}$ and so on.
Step 2:
$\int\limits_0^1e^{3-2x}dx=\lim_{h\to 0}h\{f(0)+f(0+h)+f(0+2h)+.....+f(0+(n-1)h\}$
$\qquad\qquad=\lim_{h\to 0}h\{e^2+e^{2-3h}+e^{2-6h}+....+e^{2-3(n-1)h}\}$
$\qquad\qquad=\lim_{h\to 0}he^2\{1+e^{-3h}+e^{-6h}+....+e^{-3(n-1)h}\}$
$\qquad\qquad=\lim_{h\to 0}\large\frac{he^2\{1-(e^{-3h})^n}{1-e^{-3h}}$
It is a GP series,where $a=1,r=\mid e^{-3h}\mid <1$
Step 3:
$\lim_{h\to 0}\large\frac{e^2(1-e^{-3(1))}}{1-e^{-3h}(-3)/-3h}=\large\frac{e^2(1-e^{-3})}{\lim_{h\to 0}\Large\frac{e^{-3h}-1}{-3h}}$
$\lim_{x\to 0}\large\frac{e^x-1}{x}$$=1$
$\Rightarrow \large\frac{1}{3}$$[e^2-e^{-1}]$
answered Sep 10, 2013 by sreemathi.v
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