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

1. Nov 4, 2005

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?

2. Nov 4, 2005

### Focus

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

3. Nov 4, 2005

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

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

$$\int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x) - {\frac {1} {8} x^4} = x^4(\frac {1} {2} ln (x) - \frac {1} {8})$$

Last edited: Nov 4, 2005
4. Nov 4, 2005

### mattmns

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

5. Nov 4, 2005

And then:

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

Last edited: Nov 4, 2005
6. Nov 4, 2005

### 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

7. Nov 4, 2005

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

8. Nov 4, 2005

### 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.

9. Nov 5, 2005

### HallsofIvy

Staff Emeritus
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!