# Integral of an absolute value

I am trying to evaluate the integral $$\int_{-1}^2 (x-2|x|)\,dx$$

I know that this should give me $x^2 /2 - x^2$ for the antiderivative.

I then evaluate at $2$ which gives me $2 - 4 = -2$

Then evaluate at $-1$ and get $\frac{1}{2} - 1 = -\frac{1}{2}$.

Then I find the difference $-2 - \frac{1}{2} = -2.5$

This is wrong and I do not know why.

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You should consider splitting the integral at the point where $|x|$ changes from $-x$ to $+x$, namely at $x=0$. So $$\int_{-1}^2(x-2|x|)dx = \int_{-1}^0(x-2(-x))dx + \int_0^{2}(x-2x)dx$$ $$=3\int_{-1}^0 xdx -\int_0^2xdx$$

which are integrals you should be able to evaluate.

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I dont follow what happend to the 0 to 2 (x-2x)dx part, where did the 3 go? – user138246 Apr 22 '12 at 3:15
In the range $-1 \leq x \leq 0$ we have that $|x| = -x$, so I substitute $(-x)$ where $|x|$ was in the first integral. Then $x-2(-x) = 3x$ which is where $3\int_{-1}^0xdx$ comes from. Similarly, in the range $0 \leq x \leq 2$ we have that $|x| = x$, so I replace $|x|$ by $x$ in the second integral. Then $x-2x = -x$, which is where $-\int_0^2xdx$ comes from. – nullUser Apr 22 '12 at 3:20

Hint

$$|x|=\begin{cases} -x, &x \le 0\\x. &x \ge 0\end{cases}$$

So, \begin{align}\int_{-1}^2|x|\; \mathrm{d} x&=\int_{-1}^0|x| \;\mathrm{d}x+\int_0^2|x| \;\mathrm{d}x\\&=?\end{align}

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I do not know what that really means, x will always be positive. – user138246 Apr 22 '12 at 3:12
No, @Jordan. $x$ is negative when $-1 \le x \lt 0$. This is the reason why the integral of $|x|$ is different from $x$... – user21436 Apr 22 '12 at 3:14
I thought absolute value meant that the value of that number is always positive. – user138246 Apr 22 '12 at 3:15
Will convince you on chat @Jordan – user21436 Apr 22 '12 at 3:17
|x| will be positive, yes, but the x-coordinate is negative meaning it's relation to x is -x. – TheGreatDuck Mar 8 at 17:48

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