integration - Triangle inequality for integrals proof

I'm having a bit of trouble proving the following inequality, using a graphic proof.

$$\left| \int_a^b f \right | \le \int_a^b |f|$$ It's difficult for me to imagine these sorts of problems so I'm quite lost. Thank you

Answer

Let us look at an example of $$\left| \int_a^b f \right | \le \int_a^b |f|$$

For $f(x) =x$ on the interval $[ -1,1]$ what is $\int_a^b f$ ?

The answer is $\int _{-1}^{1} xdx = x^2/2|_{-1}^{1} =0$.

What about the $|x|$ on the same interval. Now your function looks like a $V$ and it is defined as $f(x) = x$ for $x\ge 0$ and $f(x) = -x$ for $x\le 0$

Now you take the integral and you come up with $$\int _{-1}^{1}| x|dx$$ = $$\int _{-1}^{0}(- x)dx + \int _{0}^{1}( x)dx = 1$$

Clearly $0 \le 1$.

The important point is for f(x) you may have negative values but for $|f(x)|$ we only have non-negative values.