Let `g(x)=int_(0)^(x)f(t)dt`, where `f` is such that `1/2lef(t)le1`, for `tepsilon[0,1]` and `0lef(t)le1/2`, for `tepsilon[1,2]`. Then prove that `1/2leg(2)le3/2`.

Let `g(x)=int_(0)^(x)f(t)dt`, where `f` is such that `1/2lef(t)le1`, f

1.	Let `g(x)=int_(0)^(x)f(t)dt`, where `f` is such that `1/2lef(t)le1`, for `tepsilon[0,1]` and `0lef(t)le1/2`, for `tepsilon[1,2]`. Then prove that `1/2leg(2)le3/2`.
Answer» `g(x)=int_(0)^(x)f(t)dt` `:.g(2)=int_(0)^(2)f(t)dt=int_(0)^(1)f(t)dt+int_(1)^(2)f(t)dt` Now `1/2lef(t)le1` for `tepsilon[0,1]` `impliesint_(0)^(1)1/2dtleint_(0)^(1)f(t)dtleint_(0)^(1)1 dt` `implies 1/2 le int_(0)^(1)f(t)dtle1`……………….1 Also `0lef(t)le1/2` for `tepsilon[1,2]` `implies int_(1)^(2)0dt le int_(1)^(2)f(t)le int_(1)^(2)1/2dt` `implies 0 le int_(1)^(2) f(t) dt le 1/2`...............2 Adding 1 and 2 we get. `1/2 le int_(0)^(1)f(t)+int_(1)^(2)f(t)dtle3/2` `implies1/2le int_(0)^(2)f(t)dtle3/2` `implies1/2 le g(2)le3/2`

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