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.