Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Users Search
Help Dashboard Sign In Sign Up
Q&A Meta
Q&A
Posts Tags Edits
Ask Question

Post History

30%
+1 −5
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 (almost 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 (almost 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 (almost 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 (almost 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 (almost 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 (almost 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 (almost 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 (almost 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