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Multiple integral: how to retrieve abscissa range
Clash Royale CLAN TAG#URR8PPP
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We have the double integral:
$$int int_D 2x + 3y ; dx;dy$$
The domain in which we want to calculate this is the flat region defined by the curves:
$$y = x^2 ; ; ; y=x$$
Then, through the decomposition rules we resolve the internal integral to $dy$, and to do this we find the copy ordinates of the minimum and maximum points of the domain, which are precisely
$$y = x^2 ; ; ; y=x$$
While the minimum and maximum points abiscissas will be the external integral range
$$int_0^1 dx int^x_x^2 2x + 3y ; dy$$
The coordinates are found by solving to $y$ the curves that define the domain : for the abscissas, does there exist a mathematical method, or should we simply be intuitive?
Thank you in advance
integration
edited 1 hour ago
dmtri
922417
asked 3 hours ago
user3204810
1545
add a comment |Â
up vote
5
down vote
favorite
We have the double integral:
$$int int_D 2x + 3y ; dx;dy$$
The domain in which we want to calculate this is the flat region defined by the curves:
$$y = x^2 ; ; ; y=x$$
Then, through the decomposition rules we resolve the internal integral to $dy$, and to do this we find the copy ordinates of the minimum and maximum points of the domain, which are precisely
$$y = x^2 ; ; ; y=x$$
While the minimum and maximum points abiscissas will be the external integral range
$$int_0^1 dx int^x_x^2 2x + 3y ; dy$$
The coordinates are found by solving to $y$ the curves that define the domain : for the abscissas, does there exist a mathematical method, or should we simply be intuitive?
Thank you in advance
integration
edited 1 hour ago
dmtri
922417
asked 3 hours ago
user3204810
1545
2
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
We have the double integral:
$$int int_D 2x + 3y ; dx;dy$$
The domain in which we want to calculate this is the flat region defined by the curves:
$$y = x^2 ; ; ; y=x$$
Then, through the decomposition rules we resolve the internal integral to $dy$, and to do this we find the copy ordinates of the minimum and maximum points of the domain, which are precisely
$$y = x^2 ; ; ; y=x$$
While the minimum and maximum points abiscissas will be the external integral range
$$int_0^1 dx int^x_x^2 2x + 3y ; dy$$
The coordinates are found by solving to $y$ the curves that define the domain : for the abscissas, does there exist a mathematical method, or should we simply be intuitive?
Thank you in advance
integration
edited 1 hour ago
dmtri
922417
asked 3 hours ago
user3204810
1545
We have the double integral:
$$int int_D 2x + 3y ; dx;dy$$
The domain in which we want to calculate this is the flat region defined by the curves:
$$y = x^2 ; ; ; y=x$$
Then, through the decomposition rules we resolve the internal integral to $dy$, and to do this we find the copy ordinates of the minimum and maximum points of the domain, which are precisely
$$y = x^2 ; ; ; y=x$$
While the minimum and maximum points abiscissas will be the external integral range
$$int_0^1 dx int^x_x^2 2x + 3y ; dy$$
The coordinates are found by solving to $y$ the curves that define the domain : for the abscissas, does there exist a mathematical method, or should we simply be intuitive?
Thank you in advance
integration
integration
edited 1 hour ago
dmtri
922417
asked 3 hours ago
user3204810
1545
edited 1 hour ago
dmtri
922417
asked 3 hours ago
user3204810
1545
edited 1 hour ago
dmtri
922417
edited 1 hour ago
dmtri
922417
edited 1 hour ago
dmtri
922417
922417
asked 3 hours ago
user3204810
1545
asked 3 hours ago
user3204810
1545
asked 3 hours ago
user3204810
1545
1545
2
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago
add a comment |Â
2
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago
2
2
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation
$$x^2=x iff x(x-1)=0 iff x=0 lor x=1$$
Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that
$$D=(x,y)in mathbbR^2: x^2le yle x$$
answered 3 hours ago
gimusi
72.7k73888
add a comment |Â
up vote
0
down vote
You can also define new variables $$u = x, quad v = fracyx$$
We see that $u = x in [0,1]$ and $y in [x^2, x] iff v = fracyx in [x,1]=[u,1]$.
The Jacobian is given by
$$frac1J = beginvmatrix partial_xu & partial_yu \ partial_xv& partial_yvendvmatrix = beginvmatrix 1 & 0 \ -fracyx^2& frac1xendvmatrix = frac1x$$
so $|J| = |x| = x = u$.
Since $y = xv = uv$, the resulting integral is
$$int_u=0^1 int_v=u^1 (2u+3uv)u,dv,du = frac1130$$
which is the same as $int_0^1 int^x_x^2 (2x + 3y) ,dy,dx$.
answered 2 hours ago
mechanodroid
24.3k62245
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation
$$x^2=x iff x(x-1)=0 iff x=0 lor x=1$$
Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that
$$D=(x,y)in mathbbR^2: x^2le yle x$$
answered 3 hours ago
gimusi
72.7k73888
add a comment |Â
up vote
3
down vote
accepted
The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation
$$x^2=x iff x(x-1)=0 iff x=0 lor x=1$$
Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that
$$D=(x,y)in mathbbR^2: x^2le yle x$$
answered 3 hours ago
gimusi
72.7k73888
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation
$$x^2=x iff x(x-1)=0 iff x=0 lor x=1$$
Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that
$$D=(x,y)in mathbbR^2: x^2le yle x$$
answered 3 hours ago
gimusi
72.7k73888
The set up is correct and a simple plot of the graph can help to guide in order to find the abscissa range which can then be found by the equation
$$x^2=x iff x(x-1)=0 iff x=0 lor x=1$$
Note that to define the region $D$ properly we also need some other condition that is for example that the integral must be proper or that
$$D=(x,y)in mathbbR^2: x^2le yle x$$
answered 3 hours ago
gimusi
72.7k73888
answered 3 hours ago
gimusi
72.7k73888
answered 3 hours ago
gimusi
72.7k73888
answered 3 hours ago
gimusi
72.7k73888
72.7k73888
add a comment |Â
add a comment |Â
up vote
0
down vote
You can also define new variables $$u = x, quad v = fracyx$$
We see that $u = x in [0,1]$ and $y in [x^2, x] iff v = fracyx in [x,1]=[u,1]$.
The Jacobian is given by
$$frac1J = beginvmatrix partial_xu & partial_yu \ partial_xv& partial_yvendvmatrix = beginvmatrix 1 & 0 \ -fracyx^2& frac1xendvmatrix = frac1x$$
so $|J| = |x| = x = u$.
Since $y = xv = uv$, the resulting integral is
$$int_u=0^1 int_v=u^1 (2u+3uv)u,dv,du = frac1130$$
which is the same as $int_0^1 int^x_x^2 (2x + 3y) ,dy,dx$.
answered 2 hours ago
mechanodroid
24.3k62245
add a comment |Â
up vote
0
down vote
You can also define new variables $$u = x, quad v = fracyx$$
We see that $u = x in [0,1]$ and $y in [x^2, x] iff v = fracyx in [x,1]=[u,1]$.
The Jacobian is given by
$$frac1J = beginvmatrix partial_xu & partial_yu \ partial_xv& partial_yvendvmatrix = beginvmatrix 1 & 0 \ -fracyx^2& frac1xendvmatrix = frac1x$$
so $|J| = |x| = x = u$.
Since $y = xv = uv$, the resulting integral is
$$int_u=0^1 int_v=u^1 (2u+3uv)u,dv,du = frac1130$$
which is the same as $int_0^1 int^x_x^2 (2x + 3y) ,dy,dx$.
answered 2 hours ago
mechanodroid
24.3k62245
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can also define new variables $$u = x, quad v = fracyx$$
We see that $u = x in [0,1]$ and $y in [x^2, x] iff v = fracyx in [x,1]=[u,1]$.
The Jacobian is given by
$$frac1J = beginvmatrix partial_xu & partial_yu \ partial_xv& partial_yvendvmatrix = beginvmatrix 1 & 0 \ -fracyx^2& frac1xendvmatrix = frac1x$$
so $|J| = |x| = x = u$.
Since $y = xv = uv$, the resulting integral is
$$int_u=0^1 int_v=u^1 (2u+3uv)u,dv,du = frac1130$$
which is the same as $int_0^1 int^x_x^2 (2x + 3y) ,dy,dx$.
answered 2 hours ago
mechanodroid
24.3k62245
You can also define new variables $$u = x, quad v = fracyx$$
We see that $u = x in [0,1]$ and $y in [x^2, x] iff v = fracyx in [x,1]=[u,1]$.
The Jacobian is given by
$$frac1J = beginvmatrix partial_xu & partial_yu \ partial_xv& partial_yvendvmatrix = beginvmatrix 1 & 0 \ -fracyx^2& frac1xendvmatrix = frac1x$$
so $|J| = |x| = x = u$.
Since $y = xv = uv$, the resulting integral is
$$int_u=0^1 int_v=u^1 (2u+3uv)u,dv,du = frac1130$$
which is the same as $int_0^1 int^x_x^2 (2x + 3y) ,dy,dx$.
answered 2 hours ago
mechanodroid
24.3k62245
edited 2 hours ago
answered 2 hours ago
mechanodroid
24.3k62245
answered 2 hours ago
mechanodroid
24.3k62245
answered 2 hours ago
mechanodroid
24.3k62245
24.3k62245
add a comment |Â
add a comment |Â
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2
Solving $x^2=x$ gives $x=0$ and $x=1$ as margin values.
– Maam
3 hours ago