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

2 Answers
Oct 30, 2017

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)

Oct 31, 2017

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.