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

[LinkMage]

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

\int_{0}^{1} {2x^3 ln (x)} dx

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

 

[Focus]

I had an exam today and this was one of the problems:
\int_{0}^{1} {2x^3 ln (x)} dx
My answer was -1/8. Is this correct or not?

I got the integral to be...
[(2x^{4}(lnx-1)-3x^{4}/2)/4]
ln 0 is infinite, thus the integral beteween 1 and 0 is infinite?!...

 

[LinkMage]

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

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

\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})

 

[mattmns]

My calculator got -1/8, and he is smart And your work looks correct too

 

[LinkMage]

And then:

\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}

 

[mattmns]

No need for all that complicated work

You have x = 1, which means 1((1/2)ln(1) - 1/8) = -1/8
You have x= 0, which means 0(something) = 0

so the answer is -1/8

 

[LinkMage]

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

 

[mattmns]

Well, I don't know how your teacher grades, so you probably did the right thing by putting it in there. The Calculator is a TI-89.

 

[HallsofIvy]

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

And they'd be perfectly correct to take marks off if you said:
"You have x= 0, which means 0(something) = 0"

0(something) is not always 0!