Log and integrals

Mar 2009
68
0
Hi,

I need some help with ln and integrals. So I am given the equation:


x^3 ln x dx

and the answer is:

in the image below.

I have trouble with this line:

= 1/4^3 lnx - 1/4 (F) x^3

(F) being the integral sign

why is that 1/4 not in the second integral?

Hope you get what I mean.



 

Attachments

Feb 2010
1,036
386
Dirty South
Hi,

I need some help with ln and integrals. So I am given the equation:


x^3 ln x dx

and the answer is:

in the image below.

I have trouble with this line:

= 1/4^3 lnx - 1/4 (F) x^3

(F) being the integral sign

why is that 1/4 not in the second integral?

Hope you get what I mean.



You are supposed to use integration by parts here:

let \(\displaystyle u = ln(x) \) and \(\displaystyle dv = x^3 dx\)

then \(\displaystyle du = \frac{1}{x} dx\) and \(\displaystyle v = \frac{x^4}{4}\)

then calculate \(\displaystyle uv - \int v \mbox{du}\)
 
Mar 2009
68
0
Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.
 
Feb 2010
1,036
386
Dirty South
Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.
so you have: \(\displaystyle uv - \int v \mbox{du}\)

refer to the above post to see what u, v, du, and dv are.

Then \(\displaystyle uv - \int v \mbox{du} = ln(x) \times \frac{x^4}{4} - \int \frac{x^4}{4} \times \frac{1}{x} dx\)

\(\displaystyle = ln(x) \times \frac{x^4}{4} - \frac{1}{4}\int{x^4} \times \frac{1}{x} dx\)

Note: \(\displaystyle \int \frac{x}{4}\) and \(\displaystyle \frac{1}{4}\int x\) is the same.
 
Jul 2007
894
298
New Orleans
Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.

It is a constant so you can put it infront of the integral
 
Mar 2009
68
0
It is a constant so you can put it infront of the integral
Why is it not a coefficient? 1/4x^4 ? This is maybe why I got stuck ;)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
One of the things you should have learned when you started on integration, long before you got to "integration by parts", is that you can take constants out of integrals: \(\displaystyle \int c f(x) dx= c \int f(x)dx\) for c any constant.
 
Mar 2009
68
0
I did not know it was a constant and I'm still not sure why it is.
 

Defunkt

MHF Hall of Honor
Aug 2009
976
387
Israel
A constant is a number. \(\displaystyle \frac{1}{4}\) is a number, therefore you can "pull it out" of the integral.
 
Mar 2009
68
0
I think I get it now. So using the constant multiple rule, it would make no difference if that coefficient is in the integral or not, but it could be there if we wanted and the outcome would be the same.