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What is the integral of the the absolute value of x over x dx, when the lower bound is -1 and the upper bound is 2?

Hannah
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    Can you compute that $|x|/x$ is for $x > 0$ and $x < 0$, respectively? The result is pretty simple if you do so. –  Mar 10 '14 at 01:45
  • There is no value at zero just before and after it, so the integral is nonexistent? – Hannah Mar 10 '14 at 01:47
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    If a bounded function has only a single discontinuity, it's still Riemann integrable. –  Mar 10 '14 at 01:49

2 Answers2

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Yes, it's discontinuous, but you can still take the integral from -1 to 0, and 0 to 2. This is a particularly nice one to look at geometrically since you can directly calculate the area of these curves.

So I'll just note that the magnitude of this curve is -1 for $x < 0$ and +1 for $x > 0$. So you have an area of -1*1 for the first section, and +1*2 for the second section. What's the total area?

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Carser
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Remember that the function $f(x)=\displaystyle\frac{|x|}{x}$ can be written as $f(x)=1$ for $x>0$ and $f(x)=-1$ for $x<-1$. It is not defined at $0$. So we may split the integral as follows,

$\displaystyle\int_{-1}^{2}\displaystyle\frac{|x|}{x}=\displaystyle\int_{-1}^{0}-1dx+\displaystyle\int_{0}^{2}1dx$

I'll leave the evaluation for you. Note that the fact that $f(x)$ is not defined at $0$ does not affect the integral as the integral has no meaning at a single point (a bit of handwaving but should suit this particular scenario/level)

Joseph Zambrano
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