Let $\displaystyle f $be integrable in $\displaystyle [a,b]. $Prove that:
$\displaystyle
\int_{a}^{b} |f| \leq{\displaystyle\int_{a}^{b}|f|}
$
Thanks
First note that by the FTC, $\displaystyle \int_a^b f= F(b)-F(a)$.
Thus, $\displaystyle \left|\int_a^b f\right|=\left|F(b)-F(a)\right|$.
Applying the triangle inequality, we have
$\displaystyle \begin{aligned}\left|F(b)-F(a)\right| &\leq \left|F(b)\right|+\left|F(a)\right|\\ &\leq F(a)+F(b)\\ &\leq F(a)-F(x_1)+F(x_2)-F(x_1)+\ldots-F(x_{k-1})+F(b)\\ &=-(F(x_1)-F(a))+(F(x_2)-F(x_1))+\ldots+(F(b)-F(x_{k-1}))\\ &=-\int_a^{x_1}f+\int_{x_1}^{x_2}f-\int_{x_2}^{x_3}+\ldots+\int_{x_{k-1}}^bf\\ &=\int_a^b\left|f\right|\end{aligned}$
I think that's the way to approach it, but I'll let others chip in (or inform me I'm not correct)