# 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

### arildno

Glad to be of help!
To make you understand WHY it must be so, remember that "essentially", integration is to sum together a (n infinite) number of quantities, each of which quantity is the function's VALUE at some point (either positive, negative or zero) MULTIPLIED with a (n infinitely small) positve length.

Thus, although this seems to be a sum of strict areas of RECTANGLES, the HEIGHT (i.e, the function value) of those rectangles may have any type of sign.

THAT is why integration is NOT, simplistically, "the area under a curve", but rather, the net, signed "area" encompassed between a curve and the x-axis.