Define on by and
Compute If you think the function is un-integrable then input as your answer
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we have f ( x ) = 0 for x = 1 , 2 1 , 3 1 , . . . . .
f ( x ) = − 1 for 2 1 < x < 1
f ( x ) = 1 for 3 1 < x < 2 1
f ( x ) = − 1 for 4 1 < x < 3 1
and so on..............
The derived set of points of discontinuity of f is finite. Hence the function is Riemann Integrable .
So we can represent the integral of this piecewise continuous function as an infinite sum .
∫ 0 1 f ( x ) d x = − 1 ( 1 − 2 1 ) + 1 ( 2 1 − 3 1 ) − 1 ( 3 1 − 4 1 ) . . . . . . . .
= − ( 1 − 2 1 + 3 1 + . . . . . . . ) + ( 2 1 − 3 1 + 4 1 − . . . . . . . )
Using the fact that 1 − 2 1 + 3 1 − 4 1 + . . . . . = ln ( 2 )
We have our answer as 1 − 2 ln ( 2 )
PS:- I am not showing the evaluation of the limits as they are very elementary .