Does int_0^1 2x^3 ln(x)dx equal -1/8 or not

I had an exam today and this was one of the problems:

[tex] \int_{0}^{1} {2x^3 ln (x)} dx [/tex]

My answer was -1/8. Is this correct or not?

 

Here is what I did (if I remember correctly):

[tex] \int {2x^3ln(x)} dx [/tex]
[tex]u = ln (x) [/tex] [tex] v' = 2x^3dx [/tex]
[tex]u' = \frac {1} {x} [/tex] [tex] v = \frac {1} {2} x^4 [/tex]

[tex]\int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x) - {\frac {1} {8} x^4} = x^4(\frac {1} {2} ln (x) - \frac {1} {8}) [/tex]

 

And then:

[tex] \lim_{\substack{s\rightarrow 0^+}} ${\displaystyle x^4(\frac {1} {2} ln (x) - \frac {1} {8})|_{s}^{1}}$ = \lim_{\substack{s\rightarrow 0^+}} {[1^4(\frac {1} {2} ln (1) - \frac {1} {8})] - {[s^4(\frac {1} {2} ln (s) - \frac {1} {8})] = -\frac {1} {8} - 0 = -\frac {1} {8} [/tex]

 

Well, that's how they taught it. I'm sure I would be taken marks off if I didn't do it like that.

BTW, what calculator do you have? I want one like that. :P