# 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 $$\dfrac{d(|x|)}{dx} = \dfrac{d( \sqrt {x^2})}{dx} = \dfrac{x}{\sqrt {x^2}} = \dfrac{x}{|x|} \quad (II)$$

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