Integrating x and 1+X^6/L^6 from 0 to L: Seeking Help and Tips

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Discussion Overview

The discussion revolves around the integration of two expressions involving a variable x and a constant L, specifically focusing on the integrals of x(1 + X^6/L^6) and (1 + X^6/L^6) from 0 to L. Participants seek assistance in evaluating these integrals, which relate to a problem involving the center of mass of a baseball bat with a non-uniform linear density.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant requests help with the integrals of x(1 + X^6/L^6) and (1 + X^6/L^6) from 0 to L, expressing difficulty in obtaining the correct answers.
  • Another participant suggests a breakdown of the first integral into two parts, indicating it can be expressed as a sum of simpler integrals involving x and x^7.
  • A clarification is made regarding the placement of L^6 in the integrals, with one participant confirming it is only under the x^6 term.
  • A participant provides a computed result for the integral of x(1 + x^6/L^6), stating it evaluates to 5L^2/8, while noting that L is constant and the integral is straightforward.
  • Another participant raises a concern about possibly misunderstanding the problem, elaborating on the context of finding the center of mass for a baseball bat with a specific linear density.
  • A participant requests to see the work of the original poster to understand their approach and results better, while also providing guidance on using LaTeX for clarity in mathematical expressions.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the integrals, with some providing interpretations and computations while others seek clarification. There is no consensus on the correct evaluation of the integrals, and the discussion remains open-ended.

Contextual Notes

Participants have not fully resolved the assumptions regarding the integration process or the implications of the linear density function on the center of mass calculation.

PhysicsBeginner
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Hi everyone,

I need some help integrating the following. I have been trying to integrate it but can't seem to get the correct answer and what it is suppose to be. Any help would be greatly appreciated:

integral(x (1+X^6/L^6) dx) going from 0 to L

The second one is similar:
integral( 1+X^6/L^6) dx) going from 0 to L

Thanks.
 
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IF you mean

[tex]\int_0^L \frac {x(1+x^6)} {L^6} dx= \frac 1 {L^6}\int_0^L x dx + \int_0^L x^7 dx[/tex]

and

[tex]\int_0^L \frac {1+x^6} {L^6} dx[/tex]

Is very similar, can you finish?
 
Last edited:
Integral said:
Do you mean

[tex]\int_0^L \frac {x(1+x^6)} {L^6} dx[/tex]

and

[tex]\int_0^L \frac {1+x^6} {L^6} dx[/tex]

No, the L^6 is just under the x^6.
 
Ok
[tex]\int_0^L x(1 + \frac {x^6} {L^6}) dx = \int_0^L (x + \frac {x^7} {L^6}) dx = \frac {5 L^2} 8[/tex]

Since the L is constant this is really a pretty straight forward integral of a polynomial.
 
Last edited:
Perhaps I'm doing something wrong in the problem. Here is the problem and what we are suppose to do is take the integral of the numerator and denominator and get a ratio for L, where lambda = dm/dx and dm = lambda.

Once i got what dm was i substituted appropriately into the equation for the center of mass which is:

Integral (x dm) / Integral (dm)

Here is the problem:

A baseball bat of length L has a peculiar linear density (mass per unit length) given by lambda=lambda(nought) (1 + x^6/L^6). Find the x coordinate of the center of mass in terms of L.

Integral said:
Ok
[tex]\int_0^L x(1 + \frac {x^6} {L^6}) dx = \int_0^L (x + \frac {x^7} {L^6}) dx = \frac {5 L^2} 8[/tex]

Since the L is constant this is really a pretty straight forward integral of a polynomial.
 
What is the result you are getting? What are you doing to get it? I would like to see some of your work.

BTW: Click on the equation boxes I create to see the the code used to create the equations. also see the thread in General Physics with LaTex in the title (it is stickied to the top) to learn how to generate equations. This avoids confusion over what you said vs what you meant.
 

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