Is integral of a function always bigger than the function?

Question

Is $|\int_a^b f(x)dx| \ge |f(b) - f(a)|$ true?

Answer 1

No, it isn't always true. Try $f(x)=x$.

$\int_0^1 x dx = \frac{1}{2}$

but $f(1)-f(0)=1-0=1$.

Answer 2

$f(x)=-x$ for $x \in [-1,1]$

We have that $f(-1)=1$ and $f(1)=-1$ so $|f(-1)-f(1)|=|1+1|=2$

And $\int_{-1}^1 f dx = 0$.