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Grade: 12
        

integrate between 0 to infinity [3/ (x2+1)]dx, where[ ] represents greatest integer function


 

10 years ago

Answers : (2)

Akshay Ragupathy
18 Points
							When 3/(x^2 +1) < 1, greatest integer of the above = o
Therefore, for x^2 + 1 < 3 => x > 2^(1/2), [3/(x^2 +1)] = o

The integral can be written as integral of [3/(x^2 +1)] from 0 to sqrt 2.
[3/(x^2+1)] = 1 for (1/sqrt 2 < x < sqrt 2)
[3/(x^2+1)] = 2 for (0 < x < 1/sqrt2)
[3/(x^2+1)] = 3 for (x=0)

integral = (sqrt2 - (1/sqrt2)) + 2(1/sqrt2) = 3/sqrt2

						
10 years ago
Ramesh V
70 Points
							

so for 0 < x < 2-1/2   : [3/ (x2+1)] = 2

and for 2-1/2 < x < 21/2   : [3/ (x2+1)] = 1
and for 21/2 < x < infininty   : [3/ (x2+1)] = 0

so the final integral is :

 int. frm 0 to 2-1/2: (2.dx) + int. frm 2-1/2 to 21/2: (1.dx)

answer is:  3 / 21/2

10 years ago
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