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Integral Calculus
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7 Jan 2012 01:07:06 IST
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Ans is -pi/2. THING IS:THE SYMMETRY IS DESTROYED BY GREATEST INTEGER FUNCTION.if u see this let us see the graph,it is zero from x=0 to tan(inverse)1.if u see the same at pi end it is -1 from x=tan(inverse)-1 to pi.so sum is 0[tan(inverse)1-tan(inverse)0]-1[tan(inverse)1-tan(inverse)0].similarly 1[tan(inverse)2-tan(inverse)1]-2[tan(inverse)2-tan(inverse)1].henceforth it is -tan(inverse)n as n approaches infinity.hence -pi/2












The ans is 0.
as u can see..
the integration will be done in two parts due to discontinuity at pie / 2... (90 deg)
so,
from 0 to pie/2 the function is positive.........
and from pie/ 2 to pie it is negative...........
and they are symmetric about the point (pie/2 , 0)........ !!!!
so the area above the x-axis in case 1.......will be equal to area below the x-axis in case2...and thrfr, due to about signs their sum will be nil!!!