How do I integrate the discontinuous function $f$?
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
asked Nov 29 at 3:28
K.M
668312
add a comment |
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
asked Nov 29 at 3:28
K.M
668312
$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 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 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 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 at 4:20
add a comment |
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
asked Nov 29 at 3:28
K.M
668312
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
asked Nov 29 at 3:28
K.M
668312
asked Nov 29 at 3:28
K.M
668312
asked Nov 29 at 3:28
K.M
668312
asked Nov 29 at 3:28
K.M
668312
asked Nov 29 at 3:28
K.M
668312
668312
$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 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 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 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 at 4:20
add a comment |
$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 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 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 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 at 4:20
$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 at 3:59
$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 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 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 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 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 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 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 at 4:20
add a comment |
1 Answer
1
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$int_0^1 x.dx + int_1^2 (x - 1).dx + int_2^3 0.dx$
answered Nov 29 at 4:05
William Elliot
7,0702519
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 at 4:10
add a comment |
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$int_0^1 x.dx + int_1^2 (x - 1).dx + int_2^3 0.dx$
answered Nov 29 at 4:05
William Elliot
7,0702519
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 at 4:10
add a comment |
$int_0^1 x.dx + int_1^2 (x - 1).dx + int_2^3 0.dx$
answered Nov 29 at 4:05
William Elliot
7,0702519
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 at 4:10
add a comment |
$int_0^1 x.dx + int_1^2 (x - 1).dx + int_2^3 0.dx$
answered Nov 29 at 4:05
William Elliot
7,0702519
$int_0^1 x.dx + int_1^2 (x - 1).dx + int_2^3 0.dx$
answered Nov 29 at 4:05
William Elliot
7,0702519
answered Nov 29 at 4:05
William Elliot
7,0702519
answered Nov 29 at 4:05
William Elliot
7,0702519
answered Nov 29 at 4:05
William Elliot
7,0702519
7,0702519
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 at 4:10
add a comment |
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 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 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 at 4:10
add a comment |
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$f$ is a function of a single variable. Why are you using a double integral?
– helper
Nov 29 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 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 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 at 4:20