LanHikari22
  • LanHikari22
integral(ln x dx) from [0,1]. Normally. you'd get negative infinity for that, right? ( Because 0 - infinity is - infinity ), or |-inf| = inf, since some say that area should be absolute. Anyway, is there any way to play with the form as to get -1 instead?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Sorry, I misread it the first time. The value of that integral is -1, what's your question?
anonymous
  • anonymous
\[ \int_0^1 \ln(x) dx = \left[ x\ln(x)-x\right]^1_0 = -1 - 0 = -1 \]
anonymous
  • anonymous
\[\int\limits \ln (x)*1 dx=\ln x \int\limits 1 dx-\int\limits \frac{ d }{ dx }\left( \ln x \right)\int\limits 1dx=(\ln x)x-\int\limits \frac{ 1 }{ x }*x dx\] \[=x \ln x-x\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

LanHikari22
  • LanHikari22
I guess my question is why 0*(ln 0) - 0 = 0 doesn't ln 0 approach negative infinity? 0 * - inf = what? Isn't that undetermined and dependent on which is faster anyhow? Or does it really just simply reduce to zero?
jim_thompson5910
  • jim_thompson5910
@LanHikari22 the issue occurs when you reach the point `0*infinity` that's an indeterminate form (one of many) https://en.wikipedia.org/wiki/Indeterminate_form
anonymous
  • anonymous
Ah, yeah. It turns out that \(x\) approaches zero "faster" than \(\ln(x)\) approaches negative infinity, and the product goes to zero. This can be demonstrated in a number of ways, for example via L'Hospital's Rule.
jim_thompson5910
  • jim_thompson5910
you'll have to rewrite \[\Large x\ln(x)\] into \[\Large \frac{\ln(x)}{1/x}\] then use L'Hospital's rule
LanHikari22
  • LanHikari22
That just takes you into an infinite deriving loop though due to it all being indefinite
anonymous
  • anonymous
No it doesn't. Applying L'Hospital's rule gives \[\frac{1/x}{-1/x^2} =-x \] which approaches zero as \(x\rightarrow 0\).
LanHikari22
  • LanHikari22
Oh crap. For whatever reason, I never thought of cancelling, I just took myself into an infinite loop. This all makes sense now, thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.