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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral as limits of sums\[\int\limits_0^5(x+1)dx\]

$\begin{array}{1 1}\large \frac{35}{2} \\35 \\\large \frac{45}{2} \\ 45\end{array} $

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Toolbox:
  • $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]\;where\;h=\frac{b-a}{n} $
  • $\int \limits_a^b f(x)dx=\frac{(b-a)}{n}\lim_{n \to \infty} [f(a)+f(a+h)+f(a+(n-1)h)]$
Given $I=\int \limits_0^5 (x+1)dx \qquad a=0\;and\;b=5$
 
we know $h=\frac{b-a}{n}=\frac{5-0}{n}\;and\;f(x)=x+1$
 
Therefore $h=\frac{5}{n}$
 
Therefore$ \int \limits_0^5 (x+1)dx =\frac{5}{n} \lim_{n \to \infty} [f(a)+f(a+h)+....+f(a+(n-1)h)]$
 
$=\frac{5}{n} \lim_{n \to \infty} [f(0+1)+f(o+h)+....f(0+(n-1)h]$
 
$=\frac{5}{n}\lim_{n \to \infty}[(0+1)+(o+h+1)+(0+2h+1)....(0+(n-1)h+1]$
 
spliting the terms,
 
$=\frac{5}{n} \lim _{n \to \infty} [1+1+1+....nterm]+(h+2h+...(n-1)h]$
 
$=\frac{5}{n} \lim _{n \to \infty} [n+h(1+2+....(n-1)]$
 
substituting for $h=\frac{5}{h}\;and\;\sum(n-1)=\frac{n(n-1)}{2}$
 
$=\frac{5}{n} \lim _{n \to \infty}\{[n+\frac{5}{n}[\frac{n(n-1)}{2}]\}$
 
on simplifying,
 
$I=\frac{5}{n} \lim _{n \to \infty} [n+\frac{5}{2}(1-\frac{1}{n})]$
 
on applying limits,
 
$I=5+\frac{25}{2}$
 
$=\frac{35}{2}$

 

 

answered Feb 8, 2013 by meena.p
 
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