Solving Integrals for Integrating by Parts

[avr10]

Homework Statement

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

Homework Equations

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]

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]

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]

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]

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]

Hi avr10!

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

Evaluate the first part between 1 and m:

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

 

[avr10]

Hm? Why is that?

 

[HallsofIvy]

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]

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.