Mathematics

#10: Post edited by $user avatar$ Snoopy‭ · 2023-01-15T19:32:42Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
~~For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $$x\mapsto f(x,0)$$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#9: Post edited by $user avatar$ Snoopy‭ · 2023-01-08T21:38:29Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
~~For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then xxx $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#8: Post edited by $user avatar$ Snoopy‭ · 2023-01-08T21:38:14Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
~~For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then xxx $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#7: Post edited by $user avatar$ Snoopy‭ · 2023-01-08T21:34:41Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
~~For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#6: Post edited by $user avatar$ Snoopy‭ · 2023-01-08T21:33:56Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
~~For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#5: Post edited by $user avatar$ Snoopy‭ · 2023-01-05T23:56:54Z (almost 2 years ago)

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~~> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?~~
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
~~~~
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
~~~~
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
~~~~
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$
~~~~

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#4: Post edited by $user avatar$ Snoopy‭ · 2023-01-05T23:55:31Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
~~\int_a^b f(x,0)dx = \int_a^b g(x)dx~~
$$
is an "ordinary single integral" (of the function $g$).
~~Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and~~
$$
~~\int_0^1f(x,0)dx = \int_0^1 2xdx=1~~
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and

$$
\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
$$

#3: Post edited by $user avatar$ Snoopy‭ · 2023-01-05T23:52:36Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)dx = \int_a^b g(x)dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
~~\int_0^1f(x,0)dx = \int_0^1 2xdx=1\;.~~
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)dx = \int_a^b g(x)dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)dx = \int_0^1 2xdx=1
$$

#2: Post edited by $user avatar$ Snoopy‭ · 2023-01-05T23:51:20Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)dx = \int_a^b g(x)dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)dx = \int_0^1 2xdx=1\;.
$$

> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
**NO**.
For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)dx = \int_a^b g(x)dx
$$
is an "ordinary single integral" (of the function $g$).
Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)dx = \int_0^1 2xdx=1\;.
$$

#1: Initial revision by $user avatar$ Snoopy‭ · 2023-01-05T23:50:57Z (almost 2 years ago)

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> How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a  f(x) \, dx $? 

**NO**.

For any function with *two* variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $x\mapsto f(x,0)$ gives you a function in only one variable. Call this function $g(x)$. Then
$$
\int_a^b f(x,0)dx = \int_a^b g(x)dx
$$
is an "ordinary single integral" (of the function $g$). 

Consider for instance $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$
\int_0^1f(x,0)dx = \int_0^1 2xdx=1\;.
$$

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