# Integral of an absolute value function

How do I find the definite integral of an absolute value function?

For instance: $f(x) = |-2x^3 + 24x|$ from $x=1$ to $x=4$

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Find the roots (thereby splitting the function into intervals on which it doesn't change sign), and in each interval evaluate the relevant function ($+f$ or $-f$). – davin Mar 16 '12 at 15:04
To expand on @davin's comment: Use the definition of the absolute value! The absolute value equals "the inside" when "the inside" is non-negative, and equals " (-) the inside" when "the inside is negative. So you need to find where "the inside" is zero (i.e. find the roots of $-2x^3 + 24x = 0$ and possibly split the integral into two or more... – The Chaz 2.0 Mar 16 '12 at 15:09

Find the roots (thereby splitting the function into intervals on which it doesn't change sign), and in each interval evaluate the relevant function (+f or −f).

In your example, we'll take $f(x) = -2x^3+24x$, so

$$f(x) = 2x(-x^2+12) = -2x(x-\sqrt{12})(x+\sqrt{12})$$

$$\int_1^4 |f| = \int_1^{\sqrt{12}}|f| + \int_{\sqrt{12}}^4 |f| = \int_1^{\sqrt{12}} f + \int_{\sqrt{12}}^4 -f$$

I'm sure you can fill in the rest.

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I'm sorry but we know how to apply the chain rule to find the integral of $f(x)= -2x^3 +24x .$ I'm trying to figure out what gives me the derivative of the absolute value of x. aka the integral of the absolute value of x.

I was hoping that somebody could confirm the integral of the absolute value of x is the square root of $x^2.$ I feel like that's nonsense!

If only you could explain the absolute value aspect. :(

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$\lvert x\rvert=\sqrt{x^2}$. (To see why this is: try plugging in a few values, like $-3,-1,1,3$, into both sides.) – Akiva Weinberger Nov 11 '14 at 22:13