Integral Help: Solving ln(x) from 0 to 1

[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?

 

[jreelawg]

the definite integral is: integral from 0 to 1 of x*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:

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)

 

[pmqable]

Shouldn't dv=xdx?

v=x^2/2

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

 

[arildno]

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

 

[pmqable]

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

positive?

 

[arildno]

positive?

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)

 

[pmqable]

Is it?

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

 

[arildno]

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

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 .

 

[pmqable]

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 .

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 :)

 

[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.