How do I integrate the discontinuous function $f$?












0














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 $1<xle2$, since this second triangle doesn't have a point defined on $x=1$, for this discontinuous function $f$.






calculus real-analysis multivariable-calculus







share|cite|improve this question






















  • $f$ is a function of a single variable. Why are you using a double integral?
    – helper
    Nov 29 '18 at 3:59










  • 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 $x=1$.
    – K.M
    Nov 29 '18 at 4:06






  • 1




    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: math.stackexchange.com/questions/56270/…
    – mlerma54
    Nov 29 '18 at 4:11












  • 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 $(0,3)$?
    – K.M
    Nov 29 '18 at 4:20


















0














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 $1<xle2$, since this second triangle doesn't have a point defined on $x=1$, for this discontinuous function $f$.






calculus real-analysis multivariable-calculus







share|cite|improve this question






















  • $f$ is a function of a single variable. Why are you using a double integral?
    – helper
    Nov 29 '18 at 3:59










  • 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 $x=1$.
    – K.M
    Nov 29 '18 at 4:06






  • 1




    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: math.stackexchange.com/questions/56270/…
    – mlerma54
    Nov 29 '18 at 4:11












  • 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 $(0,3)$?
    – K.M
    Nov 29 '18 at 4:20
















0












0








0







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 $1<xle2$, since this second triangle doesn't have a point defined on $x=1$, for this discontinuous function $f$.






calculus real-analysis multivariable-calculus







share|cite|improve this question













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 $1<xle2$, since this second triangle doesn't have a point defined on $x=1$, for this discontinuous function $f$.






calculus real-analysis multivariable-calculus




calculus real-analysis multivariable-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 3:28









K.MK.M

686312




686312












  • $f$ is a function of a single variable. Why are you using a double integral?
    – helper
    Nov 29 '18 at 3:59










  • 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 $x=1$.
    – K.M
    Nov 29 '18 at 4:06






  • 1




    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: math.stackexchange.com/questions/56270/…
    – mlerma54
    Nov 29 '18 at 4:11












  • 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 $(0,3)$?
    – K.M
    Nov 29 '18 at 4:20




















  • $f$ is a function of a single variable. Why are you using a double integral?
    – helper
    Nov 29 '18 at 3:59










  • 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 $x=1$.
    – K.M
    Nov 29 '18 at 4:06






  • 1




    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: math.stackexchange.com/questions/56270/…
    – mlerma54
    Nov 29 '18 at 4:11












  • 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 $(0,3)$?
    – K.M
    Nov 29 '18 at 4:20


















$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 '18 at 3:59




$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 '18 at 3:59












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 $x=1$.
– K.M
Nov 29 '18 at 4:06




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 $x=1$.
– K.M
Nov 29 '18 at 4:06




1




1




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: math.stackexchange.com/questions/56270/…
– mlerma54
Nov 29 '18 at 4:11






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: math.stackexchange.com/questions/56270/…
– mlerma54
Nov 29 '18 at 4:11














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 $(0,3)$?
– K.M
Nov 29 '18 at 4:20






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 $(0,3)$?
– K.M
Nov 29 '18 at 4:20












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






share|cite|improve this answer





















  • 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 $int_1^2 (x-1) dx$.
    – K.M
    Nov 29 '18 at 4:10













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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2














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






share|cite|improve this answer





















  • 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 $int_1^2 (x-1) dx$.
    – K.M
    Nov 29 '18 at 4:10


















2














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






share|cite|improve this answer





















  • 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 $int_1^2 (x-1) dx$.
    – K.M
    Nov 29 '18 at 4:10
















2












2








2






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






share|cite|improve this answer












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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 4:05









William ElliotWilliam Elliot

7,3912720




7,3912720












  • 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 $int_1^2 (x-1) dx$.
    – K.M
    Nov 29 '18 at 4:10




















  • 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 $int_1^2 (x-1) dx$.
    – K.M
    Nov 29 '18 at 4:10


















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 $int_1^2 (x-1) dx$.
– K.M
Nov 29 '18 at 4:10






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 $int_1^2 (x-1) dx$.
– K.M
Nov 29 '18 at 4:10




















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