G(x) is the antiderivative of F(x) with G(1) = 1; Find a formula for G(x)

Click For Summary

Homework Help Overview

The discussion revolves around finding a formula for the antiderivative G(x) of a function F(x), given the condition G(1) = 1. Participants are exploring how to express G(x) in terms of F(x) while incorporating the specified value at x = 1.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the possibility of expressing G(x) as an integral of F(x) and consider the addition of a constant to satisfy the condition G(1) = 1. Questions arise regarding the necessity of knowing the value of the integral from 0 to 1 of F(x) to determine the constant.

Discussion Status

There is an ongoing exploration of different expressions for G(x), with some participants suggesting specific forms and questioning the justification of integration limits. Guidance has been offered regarding the relationship between the constants and the integration bounds, though no consensus has been reached on the final expression.

Contextual Notes

Participants note the challenge of determining the appropriate limits of integration and the implications of the condition G(1) = 1 on the formulation of G(x). There is also mention of the confusion caused by the variable used in the integral.

IntegrateMe
Messages
214
Reaction score
1
G(x) is the antiderivative of F(x) with G(1) = 1; Find a formula for G(x)...

Find a formula for G(x) in terms of F(x).

I would have just written [tex]\int_0^x{F(x)dx}[/tex] but I know the value G(1) = 1 has to be included somewhere. Can someone help me figure this one out?
 
Physics news on Phys.org


How about if you just add a constant?
 


[tex]\int_0^x{F(x)dx} + C[/tex]??
 


IntegrateMe said:
[tex]\int_0^x{F(x)dx} + C[/tex]??

Sure. If that is your G(x) then what value of C will make G(1)=1? Put x=1 and solve for C.
 


Well, wouldn't I need to know what the value of [tex]\int_0^1{F(x)dx}[/tex] is to determine that?
 


IntegrateMe said:
Well, wouldn't I need to know what the value of [tex]\int_0^1{F(x)dx}[/tex] is to determine that?

No. You can't compute it. [tex]\int_0^1{F(x)dx}[/tex] should appear as part of your answer.
 


[tex]G(x) = 1 + \int_1^xF(s)ds[/tex] ?
 


IntegrateMe said:
[tex]G(x) = 1 + \int_1^xF(s)ds[/tex] ?

Good job. Why the '?'
 


I wasn't sure of the limits of integration. They were kind of unjustified.
 
  • #10


IntegrateMe said:
I wasn't sure of the limits of integration. They were kind of unjustified.

If you'd followed the line I was suggesting you should have found [tex]C=1-\int_0^1{F(s)ds}[/tex]

Can you justify the bounds using that?
 
  • #11


The F(x) was making it confusing.
 
  • #12


IntegrateMe said:
The F(x) was making it confusing.

Maybe. If one of the limits of integration is x you shouldn't write the integral dx. Since the integration variable is a dummy you should use something besides x, like s. Makes life less confusing.
 
  • #13


Can you explain why C would allow you to justify the bounds.
 
  • #14


The integral from 0 to x of F minus the integral from 0 to 1 of F equals the integral from 1 to x of F.
 

Similar threads

Replies
2
Views
2K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K