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Evaluate the definite integral as limits of sums\[\int\limits_2^3x^2dx\]

$\begin{array}{1 1}\large\frac{19}{3} \\\large\frac{17}{3} \\ \large\frac{19}{5} \\\large\frac{23}{7} \end{array} $

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  • $\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} $
  • (ii)$ \sum (n-1)=\frac{n(n-1)}{2} \qquad \sum(n-1)^2=\frac{n(n-1)(2n-1)}{6}$
Given $ I=\int \limits_2^3 x^2dx$
Here $a=2\;and\;b=3,hence\; h=\frac{3-2}{n}=\frac{1}{n}$
$I=\int \limits_2^3 x^2dx=\frac{1}{h}\lim_{n \to \infty} [f(a)+f(a+h)+....+f(a+(n-1)h)]$
$=\frac{1}{h}\lim_{n \to \infty}[2^2+(2+h)^2+....(2+(n-1)h)^2]$
spliting the terms,
$=\frac{1}{h} \lim _{n \to \infty} [(2^2+2^2+....nterm)+2ah(1+2+3....(n-1)+h^2(1^2+2^2+3^2+.....(n-1)^2]$
(Because $(a+h)^2=a^2+2ah+h^2)$
$=\frac{1}{n} \lim _{n \to \infty} \bigg[n2^2+2 \times 2h \frac{(n(n-1)}{2}\bigg]+h^2 \bigg[\frac{n(n-1)(2n-1)}{6}\bigg]$
Because $ \sum (n-1)=\frac{n(n-1)}{2} \; and \; \sum(n-1)^2=\frac{n(n-1)(2n-1)}{6}$
on simplifying,we get
$\lim _{n \to \infty}\bigg[4+4h\frac{n-1)}{2}+h^2 \frac{(n-1)(2n-1)}{6}\bigg]$
Substituting for $h=\frac{1}{h},$
$\lim _{n \to \infty}\bigg[4+\frac{4}{h}\frac{(n-1)}{2}+h^2 \frac{(n-1)(2n-1)}{6}\bigg]$
On spliting the terms
$\lim _{n \to \infty}\bigg[4+\frac{4}{2}(1-\frac{1}{n}+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})\bigg]$
Applying the limits, $n \to \infty$
$[4+2+\frac{1}{6} \times 2] $


answered Feb 8, 2013 by meena.p
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