Solving integral with absolute value
$begingroup$
With a given $x > 0$ (I think we could restrict it to $x in [0, 3]$), I'm trying to find the following integral:
$$ int_{z = 0}^{min(x,1)} int_{y = 0}^{min(x - z, 1)} |x -z -y -1| dydz $$
However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $x < 1$, and it feels like negative when $x leq 3$, but I feel like I'm missing something.
Any hints on how should we proceed further?
calculus integration definite-integrals
asked Dec 10 '18 at 23:50
NutleNutle
332110
$endgroup$
add a comment |
$begingroup$
With a given $x > 0$ (I think we could restrict it to $x in [0, 3]$), I'm trying to find the following integral:
$$ int_{z = 0}^{min(x,1)} int_{y = 0}^{min(x - z, 1)} |x -z -y -1| dydz $$
However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $x < 1$, and it feels like negative when $x leq 3$, but I feel like I'm missing something.
Any hints on how should we proceed further?
calculus integration definite-integrals
asked Dec 10 '18 at 23:50
NutleNutle
332110
$endgroup$
add a comment |
$begingroup$
With a given $x > 0$ (I think we could restrict it to $x in [0, 3]$), I'm trying to find the following integral:
$$ int_{z = 0}^{min(x,1)} int_{y = 0}^{min(x - z, 1)} |x -z -y -1| dydz $$
However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $x < 1$, and it feels like negative when $x leq 3$, but I feel like I'm missing something.
Any hints on how should we proceed further?
calculus integration definite-integrals
asked Dec 10 '18 at 23:50
NutleNutle
332110
$endgroup$
With a given $x > 0$ (I think we could restrict it to $x in [0, 3]$), I'm trying to find the following integral:
$$ int_{z = 0}^{min(x,1)} int_{y = 0}^{min(x - z, 1)} |x -z -y -1| dydz $$
However, here I'm not sure how to deal with absolute values. By the construction, the value under the modulus seems to be almost always negative when $x < 1$, and it feels like negative when $x leq 3$, but I feel like I'm missing something.
Any hints on how should we proceed further?
calculus integration definite-integrals
calculus integration definite-integrals
asked Dec 10 '18 at 23:50
NutleNutle
332110
asked Dec 10 '18 at 23:50
NutleNutle
332110
asked Dec 10 '18 at 23:50
NutleNutle
332110
asked Dec 10 '18 at 23:50
NutleNutle
332110
asked Dec 10 '18 at 23:50
NutleNutle
332110
332110
add a comment |
add a comment |
1 Answer
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$begingroup$
Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y leq u$ and $y geq u$, we can rewrite our integral as the piecewise function:
$$ int_{0}^{min(x, 1)}left[ int_{0}^{min(u, 1)}(u-y) dy + int_{max(u, 0)}^{min(1+u, 1)}(y-u) dy right] dz $$
You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $min(a,b)$ and $max(a,b)$, along with the fact that $int_{a}^{b}f(y) dy = 0$ when $ageq b$, as seen by the fact that the region of integration $[a, b] := left{ x | aleq xleq b right}$ is empty.
After solving the inner integral for each separate region of $u$ (namely, $uleq0$, $ugeq1$, and $0leq uleq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
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1 Answer
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$begingroup$
Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y leq u$ and $y geq u$, we can rewrite our integral as the piecewise function:
$$ int_{0}^{min(x, 1)}left[ int_{0}^{min(u, 1)}(u-y) dy + int_{max(u, 0)}^{min(1+u, 1)}(y-u) dy right] dz $$
You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $min(a,b)$ and $max(a,b)$, along with the fact that $int_{a}^{b}f(y) dy = 0$ when $ageq b$, as seen by the fact that the region of integration $[a, b] := left{ x | aleq xleq b right}$ is empty.
After solving the inner integral for each separate region of $u$ (namely, $uleq0$, $ugeq1$, and $0leq uleq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
$endgroup$
add a comment |
$begingroup$
Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y leq u$ and $y geq u$, we can rewrite our integral as the piecewise function:
$$ int_{0}^{min(x, 1)}left[ int_{0}^{min(u, 1)}(u-y) dy + int_{max(u, 0)}^{min(1+u, 1)}(y-u) dy right] dz $$
You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $min(a,b)$ and $max(a,b)$, along with the fact that $int_{a}^{b}f(y) dy = 0$ when $ageq b$, as seen by the fact that the region of integration $[a, b] := left{ x | aleq xleq b right}$ is empty.
After solving the inner integral for each separate region of $u$ (namely, $uleq0$, $ugeq1$, and $0leq uleq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
$endgroup$
add a comment |
$begingroup$
Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y leq u$ and $y geq u$, we can rewrite our integral as the piecewise function:
$$ int_{0}^{min(x, 1)}left[ int_{0}^{min(u, 1)}(u-y) dy + int_{max(u, 0)}^{min(1+u, 1)}(y-u) dy right] dz $$
You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $min(a,b)$ and $max(a,b)$, along with the fact that $int_{a}^{b}f(y) dy = 0$ when $ageq b$, as seen by the fact that the region of integration $[a, b] := left{ x | aleq xleq b right}$ is empty.
After solving the inner integral for each separate region of $u$ (namely, $uleq0$, $ugeq1$, and $0leq uleq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
$endgroup$
Like you said, the hardest part here is the absolute value. We can make our lives easier by defining $u := x-z-1$, which simplifies the integrand into $|u - y|$. Since this function has a constant analytic form in the regions $y leq u$ and $y geq u$, we can rewrite our integral as the piecewise function:
$$ int_{0}^{min(x, 1)}left[ int_{0}^{min(u, 1)}(u-y) dy + int_{max(u, 0)}^{min(1+u, 1)}(y-u) dy right] dz $$
You still have to deal with the differing behavior of these integrals for different values of $u$, but this should be more mechanical than directly reasoning about the original absolute value. Just use the predictable behavior of $min(a,b)$ and $max(a,b)$, along with the fact that $int_{a}^{b}f(y) dy = 0$ when $ageq b$, as seen by the fact that the region of integration $[a, b] := left{ x | aleq xleq b right}$ is empty.
After solving the inner integral for each separate region of $u$ (namely, $uleq0$, $ugeq1$, and $0leq uleq1$), you just replace all your $u$'s by $x-z-1$ (including in your inequalities defining the different regions), and solve the outer integral. This requires more of the same case-by-case reasoning, but by this point you'll have the hang of it :)
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
answered Dec 11 '18 at 2:16
jemisjokyjemisjoky
361
361
add a comment |
add a comment |
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