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#8: Post edited by $user avatar$ trichoplax‭ · 2023-01-16T08:17:37Z (about 2 years ago)
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What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case" ?

~~1. Please see the question in the title, in reference to the para. aside my two green question marks below.~~
~~2. How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ ?~~
~~2. From $\int_{a}^{b} f (x, 0) d x$ , how exactly do you deduce $= \int_{a}^{b} f (x) d x$ ? What warrants you to drop and disregard the $, 0)$ ?~~
~~4. I disagree that $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ for these reasons.~~
~~4.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.~~
4.2. 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.
~~![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)~~

1. Please see the question in the title, in reference to the paragraph beside my two green question marks in the image below.
1. How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ ?
1. From $\int_{a}^{b} f (x, 0) d x$ , how exactly do you deduce $= \int_{a}^{b} f (x) d x$ ? What warrants you to drop and disregard the $, 0)$ ?
1. I disagree that $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ for the following reasons.
1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
1. 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.
![Pages 1132 and 1133 of Calculus Early Transcendentals](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

#7: Post edited by $user avatar$ TextKit‭ · 2023-01-01T02:28:00Z (over 2 years ago)

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~~"the line integral reduces to an ordinary single integral in this case" ?~~

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

1. Please see the para. aside my two green question marks below. What do the authors mean by _"the line integral reduces to an ordinary single integral in this case"_?
2. How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ ?
2. From $\int_{a}^{b} f (x, 0) d x$ , how exactly do you deduce $= \int_{a}^{b} f (x) d x$ ? What warrants you to drop and disregard the $, 0)$ ?
4. I disagree that $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ for these reasons.
4.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
4.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

1. Please see the question in the title, in reference to the para. aside my two green question marks below.
2. How do you symbolize _"the line integral reduces to an ordinary single integral in this case"_? $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ ?
2. From $\int_{a}^{b} f (x, 0) d x$ , how exactly do you deduce $= \int_{a}^{b} f (x) d x$ ? What warrants you to drop and disregard the $, 0)$ ?
4. I disagree that $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ for these reasons.
4.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
4.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

#6: Post edited by $user avatar$ TextKit‭ · 2023-01-01T02:27:00Z (over 2 years ago)

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Does $\int f (x, 0) d x = \int f (x) d x$ ? How?

"the line integral reduces to an ordinary single integral in this case" ?

1. Please see the para. aside my two green question marks below. What do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?
~~2. From $\int f (x, 0) d x$ , how do you deduce $= \int f (x) d x$ ? How do you drop and disregard the $\color{goldenrod}{, 0)}$?~~
3. I disagree that $\int f(x {\color{goldenrod}{, 0)}} \, dx = \int f(x) \, dx $ for these reasons.
3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
~~3.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

1. Please see the para. aside my two green question marks below. What do the authors mean by _"the line integral reduces to an ordinary single integral in this case"_?
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 $?
2. From $\int^b_a f(x \color{goldenrod}{, 0)} \, dx $, h o w e x a c t l y d o y o u d e d u c e$ = \int^b_a f(x) \, dx$? What warrants you to drop and disregard the $, 0)$ ?
4. I disagree that $\int_{a}^{b} f (x, 0) d x = \int_{a}^{b} f (x) d x$ for these reasons.
4.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
4.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

#5: Post edited by $user avatar$ TextKit‭ · 2022-12-27T01:31:48Z (over 2 years ago)

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~~1. What do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?~~
~~2. From $\int f (x, 0) d x$ , how do you deduce $= \int f (x) d x$ ? How can you simply knock or cast out the $, 0)$ part?~~
3. I disagree that $\int f (x, 0) d x = \int f (x) d x$ for these reasons.
3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
3.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

1. Please see the para. aside my two green question marks below. What do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?
2. From $\int f (x, 0) d x$ , how do you deduce $= \int f (x) d x$ ? How do you drop and disregard the $, 0)$ ?
3. I disagree that $\int f (x, 0) d x = \int f (x) d x$ for these reasons.
3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
3.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

#4: Post edited by $user avatar$ TextKit‭ · 2022-12-27T01:27:48Z (over 2 years ago)

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~~1. What exactly do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?~~
2. From $\int f (x, 0) d x$ , how do you deduce $= \int f (x) d x$ ? How can you simply knock or cast out the $, 0)$ part?
3. I disagree that $\int f (x, 0) d x = \int f (x) d x$ for these reasons.
3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
3.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

1. What do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?
2. From $\int f (x, 0) d x$ , how do you deduce $= \int f (x) d x$ ? How can you simply knock or cast out the $, 0)$ part?
3. I disagree that $\int f (x, 0) d x = \int f (x) d x$ for these reasons.
3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
3.2. 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.
![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

#3: Post edited by $user avatar$ TextKit‭ · 2022-12-27T01:27:12Z (over 2 years ago)

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~~Does $\int f(x \color{goldenrod}{, 0)} \, dx = \int f(x) \, dx $? How?~~

Does $\int f(x {\color{goldenrod}{, 0)}} \, dx = \int f(x) \, dx $? How?

#2: Post edited by $user avatar$ TextKit‭ · 2022-12-27T01:26:56Z (over 2 years ago)

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~~Does $\int f (x, 0) d x = \int f (x) d x$ ? How?~~

Does $\int f(x \color{goldenrod}{, 0)} \, dx = \int f(x) \, dx $? How?

#1: Initial revision by $user avatar$ TextKit‭ · 2022-12-27T01:26:37Z (over 2 years ago)

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Does $\int f(x, 0) \, dx = \int f(x) \, dx $? How?

1. What exactly do the authors mean by "the line integral reduces to an ordinary single integral in this case"? What I wrote in the title above?

2. From $\int f(x \color{goldenrod}{, 0)} \, dx $, how do you deduce  $= \int f(x) \, dx$? How can you simply knock or cast out the $\color{goldenrod}{, 0)}$ part? 

3. I disagree that $\int f(x {\color{goldenrod}{, 0)}} \, dx = \int f(x) \, dx $ for these reasons. 

3.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function. 

3.2. 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. 

![Image alt text](https://math.codidact.com/uploads/ldkzpx36gacjpg6irobvevgryitl)

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