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

In summary, the definite integral is: integral from 0 to 1 of ln(x). The integration by parts method was used, with u=ln(x), du=1/x dx, dv=dx, and v=x, to show that the integral is equal to [x*ln(x)] (1,0) - integral from 0 to 1 of dx. However, this resulted in an indeterminate form, so the limit was taken as x approaches 0 to find that the area is 0-1=-1, which doesn't make sense. The error was found to be in the original equation, which should have been integral of x*ln(x). This results in a negative area, as the function lies
  • #1
pmqable
13
0
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?
 
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  • #2
pmqable said:
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)
 
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  • #3
jreelawg said:
Shouldn't dv=xdx?

v=x^2/2

sorry youre right the original equation should just be integral of ln(x)
 
  • #4
And, what is the SIGN of "ln(x)" on the interval 0 to 1?
 
  • #5
arildno said:
And, what is the SIGN of "ln(x)" on the interval 0 to 1?

positive?
 
  • #6
pmqable said:
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)
 
  • #7
arildno said:
Is it?

oh wait it's negative haha... but it doesn't matter because -0=0 right?
 
  • #8
pmqable said:
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 .
 
  • #9
arildno said:
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 :)
 
  • #10
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.
 

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

1. What is the formula for solving ln(x) from 0 to 1?

The formula for solving ln(x) from 0 to 1 is ∫ln(x)dx = xln(x) - x + C, where C is the constant of integration.

2. How do I determine the limits of integration for ln(x) from 0 to 1?

The limits of integration for ln(x) from 0 to 1 are 0 and 1, as indicated by the notation "from 0 to 1". This means that you are integrating the function ln(x) between the values of 0 and 1.

3. Can I use substitution to solve this integral?

Yes, you can use substitution to solve this integral. You can let u = ln(x) and du = 1/x dx, which will result in the integral becoming ∫u du. After solving for u, you can substitute back in for ln(x) to get the final solution.

4. What is the value of ln(0)?

The value of ln(0) is undefined. This is because the natural logarithm function is not defined for x = 0. In the context of this integral, the lower limit of integration is approaching 0, but not actually reaching it.

5. Is there a shortcut method for solving this integral?

Yes, there is a shortcut method known as the "integration by parts" method. This involves breaking down the integral into two parts and using the formula ∫uv' dx = uv - ∫u'v dx to solve it. However, this method may not always be applicable and the standard method of integration should be used instead.

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