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sri valli Grade:
        

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

Answers : (2)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										
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
askIITians Faculty
2 years ago
Jitender Singh
13 Points
										
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
askIITians Faculty
2 years ago
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