$\int_0^x | t| dt = \frac{x|x|}{2}, \forall x \in \mathbb{R}$ Prove or give a counter example.
$\int_0^x | t| dt = \frac{x|x|}{2}, \forall x \in \mathbb{R}$
so, if $x < 0 \Rightarrow |t| = -t \Rightarrow\int_0^x (-t) dt = \frac{(-x)x}{2}$
for $x \ge 0 \Rightarrow |t| = t \Rightarrow\int_0^x t dt = \frac{xx}{2}$
but this doesn't seems right .. i think is missing the "proof". or for $\epsilon > 0$ that integral became different of the $\frac{x|x|}{2}$ because of the small difference of upper sums and lower sums .
well, i don't know .. thanks for any help

*

*$\forall x>0$ , $\displaystyle \int_{0}^x |t|$ $\mathrm{d}t=\displaystyle \int_{0}^x t$ $\mathrm{d}t=\dfrac{x^2}{2}$

*$\forall x<0$ , $\displaystyle \int_{0}^x |t|$ $\mathrm{d}t=-\displaystyle \int_{x}^0 (-t)$ $\mathrm{d}t=-\dfrac{x^2}{2}$

*$x=0$ $\implies$  $\displaystyle \int_{0}^x |t|$ $\mathrm{d}t=0$
$\therefore$ $\displaystyle \int_{0}^x |t|$ $\mathrm{d}t=\dfrac{x|x|}{2}$
