# Integral of the absolute Value of $x$: $\int |x| dx$

 

Evaluate the integral $\int |x| \; dx$ Rewrite as $\int |x| \; dx = \int 1 \cdot |x| \; dx \quad (I)$
Note that $|x| = \sqrt {x^2}$ and hence $\dfrac{d(|x|)}{dx} = \dfrac{x}{\sqrt {x^2}} = \dfrac{x}{|x|} \quad (II)$

Apply the
integration by parts : $\displaystyle \int u' v \; dx = u v - \int u v' \; dx$ to the integral on the right side of (I) above.
Let $u' = 1$ , $v = |x|$ which gives $u = x$ and $v' = \dfrac{x}{|x|}$ , see (II) above.

We substitute all the above in the formula of the integration by parts given above.
$= x |x| - \int x \dfrac{x}{|x|} dx \quad (III)$ Multiply the numerator and denominator of the integrand $x \dfrac{x}{|x|}$ in the above integral $x \dfrac{x}{|x|} = x \dfrac{x}{|x|} \dfrac{|x|}{|x|}$ Simplify noting that $|x| |x| = x^2$ $= |x|$ We now substitute the integrand $x \dfrac{x}{|x|}$ in (III) by $|x|$ and write $\int |x| dx = x |x| - \int |x| dx$ Add $\displaystyle \int |x| dx$ to both sides $\int |x| dx + \int |x| dx = x |x| - \int |x| dx + \int |x| dx$ Simplify by grouping $2 \int |x| \; dx = x |x|$ The final answer is given by $\boxed {\int |x| \; dx = \dfrac{1}{2} x |x| }$