Question

How to prove or disprove ? if `f` is integrable on `[a,b]` then `int_a^b|f(x)|dx<=|int_a^bf(x)dx|`

Answer 1

Consider `f(x)=x`, and `[a,b] = [-1,1]`

Explanation:

In fact any non-constant odd function `f` on `[-c,c]` will have

`int_-c^c abs(f(x))dx > 0 = abs(int_-c^c f(x) dx)`

Answer 2

Use Riemann definition for integrals:
`int_a^bf(x)\ dx=lim_(N->oo)sum_(i=0)^Nf(a+(b-a)i/N)/N`.

We need to find how `int_a^b|f(x)|\ dx` compares with `|int_a^bf(x)\ dx|`.

Convert both to limit and summation form using the Riemann definition for the integrals above:
`{(lim_(N->oo)sum_(i=0)^N|f(a+(b-a)i/N)|/N),(|lim_(N->oo)sum_(i=0)^Nf(a+(b-a)i/N)/N|=lim_(N->oo)|sum_(i=0)^Nf(a+(b-a)i/N)/N|):}`

There is an identity stating that `|a|+|b|+|c|+ldots≥|a+b+c+ldots|`. Thus, it must be the case that `lim_(N->oo)sum_(i=0)^N|f(a+(b-a)i/N)|/N≥lim_(N->oo)|sum_(i=0)^Nf(a+(b-a)i/N)/N|`.

The statement in the question is false.