Here you have to break the integrals.
When 1<x<2 ,[x]=1
When 2<x<3 ,[x]=2 and so on
So the given question=
[1 ]
[2] f '(x)dx+
[2][3]2*f '(x)+.....+
[[a] ][a][a]*
f '(x)dx
=f(2)-f(1)+2(f(3)-f(2))+......+[a](f(a)-f([a]))
=[a]f(a)-{f(1)+f(2)+.....+f([a])}
So answer is 4)
Moderation Team
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