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#10: Post edited by user avatar Snoopy‭ · 2023-01-15T19:32:42Z (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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
  • <p>
  • $$
  • \int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
  • $$
  • </p>
  • is an "ordinary single integral" (of the function $g$).
  • Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
  • <p>
  • $$
  • \int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
  • $$
  • </p>
  • > 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 (about 1 year ago)
  • > 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
  • <p>
  • $$
  • \int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx
  • $$
  • </p>
  • is an "ordinary single integral" (of the function $g$).
  • Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
  • <p>
  • $$
  • \int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1
  • $$
  • </p>
#3: Post edited by user avatar Snoopy‭ · 2023-01-05T23:52:36Z (about 1 year ago)
  • > 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 (about 1 year ago)
  • > 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 (about 1 year ago)
> 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
$$
\int_0^1f(x,0)dx = \int_0^1 2xdx=1\;.
$$