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Let $f(x) = x$ for $0 \le x \le 1$, $x-1$ for $1 < x \le 2$, $0$ for $2<x \le 3$.

I ended up writing the double integral as $\iint_{D_3 \cup D_4} f = \int_0^2 dx \int_0^1 2x dy = 2 \int_0^1 dx \int_0^1 x dy$. For $2 \int_0^1 dx \int_0^1 x dy$, it feels like I'm adding the triangle defined on $0 \le x \le 1$ twice. I wanted to know how I'm able to do this for this specific function since $f(x)$ is not continuous. If $g$ is continuous on an interval, then $g$ has an antiderivative on $I$. But since the converse doesn't always hold true, it would be nice if someone could explain how we're able to integrate the triangle defined on $1<x\le2$, since this second triangle doesn't have a point defined on $x=1$, for this discontinuous function $f$.

Skm
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  • $f$ is a function of a single variable. Why are you using a double integral? – ATOMP Nov 29 '18 at 03:59
  • I saw an example in my textbook, and decided to separate the region $D$ into two triangles $D_3$ and $D_4$, and integrate over the region $D_3 \cup D_4$, since triangle $D_4$ is not defined at $x=1$. – Skm Nov 29 '18 at 04:06
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    Not sure if what is bothering you is the fact that the second interval $(1,2]$ is not closed. Integration over a not closed interval has been discussed e.g. here: https://math.stackexchange.com/questions/56270/which-integral-limits-to-choose-when-dealing-with-open-intervals – mlerma54 Nov 29 '18 at 04:11
  • yes, that's what was bothering me. I also was wondering if the initial integral I did with the two variables is incorrect, or is it just superfluous given that we're considering $x$ in the interval $(0,3)$? – Skm Nov 29 '18 at 04:20

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$\int_0^1 x.dx + \int_1^2 (x - 1).dx + \int_2^3 0.dx$

William Elliot
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  • I thought about doing that at first, but for $\int_1^2 (x-1) dx$ we have that the function $x-1$ is not defined at $1$, so I'm don't know why I'm still able to integrate $\int_1^2 (x-1) dx$. – Skm Nov 29 '18 at 04:10