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 \[ \boxed {\int |x| \; dx = \dfrac{1}{2} x |x| } \]

More References and Links

1. Table of Integral Formulas
2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
   
3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
   
4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8