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Integral Calculus
Comments (3)
Firstly, since the function is symmetric about pi thrfr the integarte from {0, pi} will be +ve of that from {pi, 2pi} which will be -ve as the graph will be below X-axis for {pi, 2pi} and above X-axis for {0, pi thrfr, the ans is 0.
To solve questions relates to [ ] always go for graphical approach
In this case,
0< f(x) < 1 for {0, pi/6}.....................thrfr, [ f(x) ] = 0
1<= f(x) < 2 for { pi/6 , pi/2}...............thrfr, [ f(x) ] = 1
f(x) = 2 for pi/2 only...........thrfr, [ f(x) ] = 2 but
since, this is discontinuity so it should be taken separately but since it is just a point and further for {pi/2 , 5pi/6} the graph is again 1<= f(x) < 2 thrfr we can directly integrate it from {pi/6 , 5pi/6}.
and from {5pi/6, pi}..................0< f(x) < 1 ...............thrfr, [ f(x) ] = 0
and similarly for further {pi, 2pi} divide it as required and u'll see the graph repeats but BELOW the X=axis!!!!
thrfr the sum total 0!!!!!!!!!












-2[cosx]=-2[1-1]=0