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# What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case" ?

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1. Please see the question in the title, in reference to the paragraph beside my two green question marks in the image below.
2. How do you symbolize "the line integral reduces to an ordinary single integral in this case"? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx$?
3. From $\int^b_a f(x \color{goldenrod}{, 0)} \, dx$, how exactly do you deduce $= \int^b_a f(x) \, dx$? What warrants you to drop and disregard the $\color{goldenrod}{, 0)}$?
4. I disagree that $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx$ for the following reasons.
1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
2. The left side requires you to evaluate $f(x, y)$ at $y = 0$. $f(x)$ requires no evaluation!

I scanned James Stewart, Daniel Clegg, Saleem Watson's Calculus Early Transcendentals, 9 edn 2021, pp. 1132-3.

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wrong interpretation of the phrase in 2 (1 comment)
Title equation doesn't appear in excerpt (2 comments)

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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

$$\int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1$$
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I don't understand the context of the "no" (3 comments)
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A line integral is integrated with respect to arc parameter. If the path you're integrating along is the $x$-axis, then the arc parameter can be taken to be just $x$, and so the integral is identical to an ordinary integral with respect to $x$.

You are right that it's a bit improper to write $f(x,0) = f(x)$, since the left-hand side is a two variable function with one variable evaluated at $y=0$, while the right-hand side is just a single variable function (though it may still be useful if you can tolerate some sloppiness).

However I don't see anywhere that Stewart wrote such an equation.

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