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-   -   Integrating by parts? (http://www.physicsforums.com/showthread.php?t=250603)

 avr10 Aug18-08 02:16 AM

Integrating by parts?

1. The problem statement, all variables and given/known data

$$\text {Evaluate } \int^m_1 x^{3}ln{x}\,dx$$

2. Relevant equations

3. The attempt at a solution

Integrating by parts, but not sure which term to substitute out...it's not turning out clean...argh I've done every other problem except for this one, can someone just provide the first step? Much appreciated.

 cristo Aug18-08 02:19 AM

Re: Integrating by parts?

I don't know what you mean by "I'm not sure which term to substitute out?" What is the method for integration by parts: you should have a formula, no? What specifically are you confused with, with respect to this formula?

 Defennder Aug18-08 02:23 AM

Re: Integrating by parts?

You should try substituting for v and du, the terms for which vdu is easy to integrate. It's harder to integrate ln than it is to differentiate it, so that provides an obvious choice.

 HallsofIvy Aug18-08 04:51 AM

Re: Integrating by parts?

Since you have x3ln x dx, and want u(x) dv, there are really just two choices: either u(x)= x3 and dv= ln(x)dx or u(x)= ln(x) and dv= x3dv. Try both and see which gives you a decent integral!

 avr10 Aug18-08 04:55 AM

Re: Integrating by parts?

Thanks, I integrated it successfully, but since this is a definite integral, at the end I'm not quite sure what to do with the boundaries, namely with the m term. Is it okay to leave the answer as an expression of both m and x?

It comes out to this:

$$\frac {1}{4} (x^{4} ln{x} - \int^m_1 x^{3} \,dx)$$

 tiny-tim Aug18-08 05:29 AM

Hi avr10! :smile:
Quote:
 Quote by avr10 (Post 1838791) $$\frac {1}{4} (x^{4} ln{x} - \int^m_1 x^{3} \,dx)$$
Evaluate the first part between 1 and m:

$$\left[\frac {1}{4} x^{4} ln{x} \right]^m_1$$ :smile:

 avr10 Aug18-08 05:32 AM

Re: Integrating by parts?

Hm? Why is that?

 HallsofIvy Aug18-08 05:37 AM

Re: Integrating by parts?

Because that's what the little numbers at top and bottom of the integral sign mean!
$$\int_a^b f(x)dx= F(b)- F(a)$$
where F is an anti-derivative of f. The result of a definite integral is a number, not a function of x.

 Tomtom Aug18-08 06:47 AM

Re: Integrating by parts?

Integrating by parts using the standard formula gives me
$$\left[ln(x)*\frac {1}{3} x^{3} -\frac {1}{9}x^{3}\right]^m_1$$

Following up by the formula presented by HallsofIvy, I get:

$$ln(m)-\frac{3}{9}*m^{3}-\frac{1}{9}*m^{3}+\frac{1}{9}$$

Which is pulled together to:

$$ln(m)+\frac {1}{9}-\frac {4}{9}m^3$$

Voila?

Edit: HallsofIvy, in this case the answer is not a number, but yes, it is a constant.

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