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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 =
7 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans: 1Sol:$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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago
Jitender Singh
13 Points
										Ans: 0There 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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago
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