Quick Question Checking an example where inequality is strict I posted this earlier and there is a question after this where I give an example of a function so that the inequality is strict. Basically my understanding is that they're looking for a function that is negative in a subinterval. 
So, $f(x)=x^2-1$ in the interval $[-2,2]$ would have a strict inequality:
$$\left|\int_a^bf(x)dx\right|<\int_a^b|f(x)|dx$$
Does this work? Does anybody have a "better" example that could enhance my understanding of this question perhaps? :) I'm just solving these to review/prepare for a midterm anyways
 A: just take $f(x) = -1$ on $[-1,0]$ and $f=1$ on $(0,1]$ ! 
if you're looking for a continuous example (which is not necessary for the inequality to hold) then yours is just fine.
note that this is the continuous analogue of the triangular inequality $|a+b| \leq |a| + |b|$
EDIT : if you are looking for an example which is true for ALL a and b then you'll have to work with some pretty complicated function. Let's work with a function defined on [0,1] (you can extend it to $R$ by periodicity if you want to).
Choose a Cantor set $C$ which has positive but not full measure on [0,1] (look up Cantor space if you don't know what it is : roughly speaking, it is obtained by removing an interval centered at the middle of [0,1] from [0,1], then removing another interval centered at the middle of those 2 intervals, etc. and by choosing carefully the lengths of the interval removed).
Set $f(x) = 1$ if $x \in C$ and -1 otherwise. ($f$ is horrible)
Then for any $a$, $b$ in [0,1 the inequality will be strict as there will be a subset X of [a,b] such that $\int_X f <0$ and $\int_{[a,b]\backslash X} f>0$ wich is essentially what you need (compensation in the left term of the inequality). 
