Question

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 1

`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`