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
Jim H
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)#

NickTheTurtle
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.

Related questions
Impact of this question
1697 views around the world
You can reuse this answer
Creative Commons License