Integration help

[PhysicsBeginner]

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.

[Integral]

IF you mean

\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

and

\int_0^L \frac {1+x^6} {L^6} dx

Is very similar, can you finish?

[PhysicsBeginner]

Do you mean

\int_0^L \frac {x(1+x^6)} {L^6} dx

and

\int_0^L \frac {1+x^6} {L^6} dx

No, the L^6 is just under the x^6.

[Integral]

Ok
\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

Since the L is constant this is really a pretty straight forward integral of a polynomial.

[PhysicsBeginner]

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.

Ok
\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

Since the L is constant this is really a pretty straight forward integral of a polynomial.

[Integral]

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.