How can I change the functions in the top and bottom of a double integral?
$begingroup$
How can I go from an integral looking like this:
$int_{0}^{2}int_{frac{y}{2}}^{1} f(x,y) dxdy $
To an integral looking like this:
$int_{0}^{1}int_{0}^{2x} f(x,y) dxdy $
I just want to know what the process is behind it and how we can change the integrals around?
calculus
$endgroup$
add a comment |
$begingroup$
How can I go from an integral looking like this:
$int_{0}^{2}int_{frac{y}{2}}^{1} f(x,y) dxdy $
To an integral looking like this:
$int_{0}^{1}int_{0}^{2x} f(x,y) dxdy $
I just want to know what the process is behind it and how we can change the integrals around?
calculus
$endgroup$
add a comment |
$begingroup$
How can I go from an integral looking like this:
$int_{0}^{2}int_{frac{y}{2}}^{1} f(x,y) dxdy $
To an integral looking like this:
$int_{0}^{1}int_{0}^{2x} f(x,y) dxdy $
I just want to know what the process is behind it and how we can change the integrals around?
calculus
$endgroup$
How can I go from an integral looking like this:
$int_{0}^{2}int_{frac{y}{2}}^{1} f(x,y) dxdy $
To an integral looking like this:
$int_{0}^{1}int_{0}^{2x} f(x,y) dxdy $
I just want to know what the process is behind it and how we can change the integrals around?
calculus
calculus
asked Jan 27 at 1:50


George HarrisonGeorge Harrison
82210
82210
add a comment |
add a comment |
1 Answer
1
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$begingroup$
To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $dfrac y2$ to $1$ and $y$ from $0$ to $2$. Sketch these two curves to visualize it.
You should consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$.
Then you get $int_0^1int_0^{2x}f(x,y)dxdy$
$endgroup$
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $dfrac y2$ to $1$ and $y$ from $0$ to $2$. Sketch these two curves to visualize it.
You should consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$.
Then you get $int_0^1int_0^{2x}f(x,y)dxdy$
$endgroup$
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
add a comment |
$begingroup$
To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $dfrac y2$ to $1$ and $y$ from $0$ to $2$. Sketch these two curves to visualize it.
You should consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$.
Then you get $int_0^1int_0^{2x}f(x,y)dxdy$
$endgroup$
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
add a comment |
$begingroup$
To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $dfrac y2$ to $1$ and $y$ from $0$ to $2$. Sketch these two curves to visualize it.
You should consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$.
Then you get $int_0^1int_0^{2x}f(x,y)dxdy$
$endgroup$
To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $dfrac y2$ to $1$ and $y$ from $0$ to $2$. Sketch these two curves to visualize it.
You should consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$.
Then you get $int_0^1int_0^{2x}f(x,y)dxdy$
answered Jan 27 at 2:06
Key FlexKey Flex
8,63561233
8,63561233
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
add a comment |
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
$begingroup$
@George Harrison If you find my answer helpful then accept the answer $checkmark$ :)
$endgroup$
– Key Flex
Feb 17 at 21:57
add a comment |
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