# Definite integral help

1. Mar 11, 2012

### pmqable

the definite integral is: integral from 0 to 1 of ln(x). i used integration by parts (u=ln(x), du=1/x dx, dv=dx, v=x) to show that the integral is equal to:

[x*ln(x)] (1,0) - integral from 0 to 1 of dx.

this gives 1*ln(1)-0*ln(0)-(1-0)

ln(1)=0, so the equation is now

0*ln(0)-1

0*ln(0) is an indeterminate form, so i used the limit:
lim x-->0 x*ln(x)
lim x-->0 ln(x)/(1/x)
lim x-->0 (1/x)/(-1/x^2)
lim x-->0 -x
=0.

so the area is 0-1=-1, which doesn't make sense. what did i do wrong?

Last edited: Mar 11, 2012
2. Mar 11, 2012

### jreelawg

Shouldn't dv=xdx?

v=x^2/2

uv-integral(vdu)=(x^2lnx)/2-integral((x/2)dx)

=[(x^2lnx)/2-x^2/4] from 0 to 1.

ln(1)/2-1/4
=0-1/4
=-(1/4)

Last edited: Mar 11, 2012
3. Mar 11, 2012

### pmqable

sorry youre right the original equation should just be integral of ln(x)

4. Mar 11, 2012

### arildno

And, what is the SIGN of "ln(x)" on the interval 0 to 1?

5. Mar 11, 2012

### pmqable

positive?

6. Mar 11, 2012

### arildno

Is it?
(I wrote "ln(x)", I didn't mean the absolute value of ln(x), just ln(x), but I used citation marks. Sorry)

7. Mar 11, 2012

### pmqable

oh wait it's negative haha... but it doesn't matter because -0=0 right?

8. Mar 11, 2012

### arildno

I don't get what you say.
You get a negative area, because the area lies fully BENEATH the x-axis.

Integration yields you the net, signed area betwen a curve, and the x-axis .

9. Mar 11, 2012

### pmqable

ok thank you very much... believe it or not, every integral i have ever done has resulted in a positive area. i just integrated x^2-4 from -1 to 1 and and sure enough, i got a negative area. lesson learned :)

10. Mar 12, 2012