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# if[x] denotes the greatest integer less than or equal to x then integral of [1/1+x2]dx in limits from 1 to infinity =

Jitender Singh IIT Delhi
6 years ago
Ans: 1
Sol:
$I = \int_{1}^{\infty }[\frac{1}{1+x^{2}}]dx$
$I = \int_{1}^{2}[\frac{1}{1+x^{2}}]dx + \int_{2}^{3}[\frac{1}{1+x^{2}}]dx + .....$
$1\leq x\leq 2\Rightarrow [\frac{1}{1+x^{2}}] = 1$
$2\leq x\leq 3\Rightarrow [\frac{1}{1+x^{2}}] = 0$
Similarly zero for the next intervals.
$I = \int_{1}^{2}dx = 1$
Cheers!
Thanks & Regards
Jitender Singh
IIT Delhi
Jitender Singh
13 Points
6 years ago
Ans: 0
There is one minor mistake in the above solution. Sorry about that. When x varies from 1 to 2, then integral part of the 1/(1+x2) should be equal to zero. So the integral value should be zero.

Thanks & Regards
Jitender Singh
IIT Delhi