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

2 answers  ·  posted 2y ago by TextKit‭  ·  edited 2y ago by trichoplax‭

Question line-integral
#8: Post edited by user avatar trichoplax‭ · 2023-01-16T08:17:37Z (over 2 years ago)
Tidying 
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^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 $, 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 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^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?
  • 1. 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 $\color{goldenrod}{, 0)}$?
  • 1. 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.
  • 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)
  • "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^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 $, 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 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^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 $, 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 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)
  • Does $\int f(x {\color{goldenrod}{, 0)}} \, dx = \int f(x) \, dx $? 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 \color{goldenrod}{, 0)} \, dx $, how do you deduce $= \int f(x) \, dx$? 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 $, 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 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)
  • 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 \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)
  • 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 \color{goldenrod}{, 0)} \, dx $, how do you deduce $= \int f(x) \, dx$? 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)
#4: Post edited by user avatar TextKit‭ · 2022-12-27T01:27:48Z (over 2 years ago)
  • 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)
  • 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 \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)
#3: Post edited by user avatar TextKit‭ · 2022-12-27T01:27:12Z (over 2 years ago)
  • 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)
  • Does $\int f(x, 0) \, dx = \int f(x) \, dx $? 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) 
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)
line-integral