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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral as limits of sums\[\int\limits_a^bx\;dx\]

$\begin{array}{1 1}\large \frac{b^2-a^2}{2} \\ \large \frac{b^2+a^2}{2} \\ \large \frac{a^2-b^2}{2} \\ \large \frac{b^2-a^2}{4} \end{array} $

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1 Answer

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Toolbox:
  • $\int\limits_a^b f(x)dx=lim_{h->0}[f(a)+f(a+h)+...f(a+(n-1))]\;when\;h=\frac{b-a}{n} $
  • $\int \limits_a^b f(x)dx=(b-a)\lim_{n \to \infty} \frac{1}{n}[f(a)+f(a+h)+f(a+(n-1)h)]$
  • (ii)$ \sum (n-1)=\frac{n(n-1)}{2}$
Given $I=\int \limits_a^b xdx$
 
we know $h=\frac{b-a}{n}$
 
Therefore \int \limits_a^b xdx =(b-a) \lim_{n \to \infty \frac{1}{n}[f(a)+f(a+h)+....+f(a+(n-1)h)]$
 
After spliting a and h seperately we get
 
$=(b-a) \lim _{n \to \infty} \frac{1}{n}[a+a+h+....+a+(n-1)h)]$
 
$=(b-a) \lim _{n \to \infty} \frac{1}{n}[a+a+a+...nterm)+\frac{(b-a)}{n}(1+2+3+....(n-1)]$
 
Since (a+a+a.....nterm)=na
 
and $1+2+3+......(n-1)=\frac{n(n-1)}{2}$
 
and $h=\frac{(b-a)}{n}$
 
we can substitute n a and $\frac{n(n-1)}{2}$
 
Therefore $(b-a) \lim _{n \to \infty} \frac{1}{n}[na+\frac{(b-a)}{n}\frac{(n)(n-1)}{2}]$
 
$=(b-a) \lim _{n \to \infty} a +\frac{(n-1)}{2n}(b-a)]$
 
$=(b-a) \lim _{n \to \infty} a +(\frac {1}{2}-\frac{1}{n})(b-a)]$
 
applying the limits we get ,
 
$(b-a)[a+\frac{1}{2}(b-a)]$
 
$=\Large\frac{(b-a)(b+a)}{2}=\frac{b^2-a^2}{2}$

 

 

answered Feb 8, 2013 by meena.p
 
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