Solve for y: F(3) = 1/4; Integration: f(3) = 1/4; y = 1/(6x-x^2+13)

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Homework Help Overview

The discussion revolves around solving for y in the context of an integration problem involving the function y = 1/(6x - x^2 + 13) and the condition f(3) = 1/4. Participants are analyzing the integration steps and constants involved in the derivation of y.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of dy/y^2 and its implications on the derived equation for y. There are questions about the distribution of the negative sign and the constant C in the integration process. Some participants express confusion regarding the correctness of the derived answer and the relationship between C and the given conditions.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants have offered guidance on the integration steps, while others have questioned the assumptions made regarding the constants and the signs in the equations. There is no explicit consensus on the correctness of the answers presented.

Contextual Notes

Participants are working under the constraints of a homework problem, which may involve specific rules or expectations regarding the presentation of solutions. There is also mention of a reference answer from a book that is being compared against the participants' calculations.

UrbanXrisis
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f(3)=1/4

[tex]\int\frac{dy}{y}=\int (6-2x)dx[/tex]
[tex]-\frac{1}{y}=6x-x^2+C[/tex]
[tex]-4=18-9+C[/tex]
[tex]C=-13[/tex]
[tex]-\frac{1}{y}=6x-x^2-13[/tex]
[tex]y=-\frac{1}{6x-x^2-13}[/tex]
the answer is...
[tex]y=\frac{1}{6x-x^2+13}[/tex]

why?
 
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Looks like the answer is wrong.
 
I think that the negative was distributed before the Constant was added.. I just don't know why?
 
[tex]\int \frac{dy}{y}[/tex] is not [tex]\frac{-1}{y}[/tex]
 
sorry, it's y^2

f(3)=1/4

[tex]\int\frac{dy}{y^2}=\int (6-2x)dx[/tex]
[tex]-\frac{1}{y}=6x-x^2+C[/tex]
[tex]-4=18-9+C[/tex]
[tex]C=-13[/tex]
[tex]-\frac{1}{y}=6x-x^2-13[/tex]
[tex]y=-\frac{1}{6x-x^2-13}[/tex]
the answer is...
[tex]y=\frac{1}{6x-x^2+13}[/tex]

why?
 
[tex]-\frac{1}{y}=6x-x^2+C[/tex]

[tex]y = -\frac{1}{6x-x^2+c}[/tex]

[tex]-\frac{1}{4} = \frac{1}{9+C}[/tex]

C = 13
 
sub that back in... you get the same equation as I do...
 
I'm sorry, C is -13, and yeah, the one you got is correct.
 
the book's answer is http://home.earthlink.net/~urban-xrisis/phy001.jpg

I don't see how they got that answer...
 
Last edited by a moderator:
  • #10
You wrote the answer in correct in your first post, and unless there's a new math system where

C + 9 = 4, and C = 18 theyre wrong. C is -13
 
  • #11
I took the answers right off of the college-board website! wow...
 
  • #12
Nobody's infall-yable.
 

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