The function `f` and `g` are positive and continuous. If `f` is increasing and `g` is decreasing, then `int_0^1f(x)[g(x)-g(1-x)]dx` is always non-positive is always non-negative can take positive and negative values none of these

Options

A. is always non-positive
B. is always non-negative
C. can take positive and negative values
D. none of these

Answer

A

Explanation

`I=int_(0)^(1)f(x)[g(x)-g(1-x)]dx`
`=-int_(0)^(1)f(1-x)[g(x)-g(1-x)]dx`
or `2I=int_(0)^(1)[f(x)-f(1-x)][g(x)-g(1-x)]dx`
`=2int_(0)^(1//2)[f(x)-f(1-x)].[g(x)-g(1-x)]dxle0`