[SPLIT] int [1,e] [x^2 ln(x)] dx (please check)

warwick

Full Member
Joined
Jan 27, 2006
Messages
311
As far as I can tell, I think I'm doing this correctly:

integral of x^2 lnx dx from 1 to e

u = ln x, u' = 1/x, v = (1/3)x^3, v' = x^2 dx

= ln x (1/3)x^3 - (1/6)x^2

= (2/6)e^3 - (1/6)e^2 + (1/6)
 
You're on the right track, except 3*3=9, not 6. I know, I know, those are the little things we do that make all the difference.

\(\displaystyle \L\\\frac{x^{3}ln(x)}{3}-\frac{1}{3}\int{x^{2}}dx\)

\(\displaystyle \L\\\frac{x^{3}ln(x)}{3}-\frac{x^{3}}{9}=\frac{x^{3}(3ln(x)-1)}{9}\)

Now, use the limits of integration:

\(\displaystyle \L\\\left[\frac{e^{3}(3ln(e)-1)}{9}\right]-\left[\frac{3ln(1)-1}{9}\right]=\frac{2e^{3}}{9}+\frac{1}{9}\approx{4.57456}\)

Just a thought, but since you're a regular, warwick, perhaps learn a little LaTex. It'll make your posts look nicer. Start out small at first. You can check the code I used by clicking on 'quote' at the upper right corner of my posts.
 
Yeah, I thought about that. I'll look into.

Also, I just found my typographical error. I had a square where there should have been a cube.
 
Sure!

1) If you make it worse, you may wish to try something else.
2) If it's cyclical, try it for the cycle before you give up.
3) If you can change from transcendental, trigonometric, or worse to algebraic, that's probably a good idea.

In other words, no, not really. galactus has it right "practice and observation". I might add perseverence and consistency.
 
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