Integrals

[Aki]

I have the function f(x)=1/x , and there's two asymptotes, the x-and y-axes. As x gets larger, the f(x) becomes smaller, but it is never 0. So my question is, what is the intergral from x=1 to x=infinite?

[Sirus]

\int \frac{1}{u}\,du=\ln{|u|}+C

[digink]

The derivative of ln x is 1/x so like the person stated above

\int (1/u)du = ln|u| + C

[Aki]

I'm sorry, but I'm so lost already. Could somebody please explain it to me? What is u?

[digink]

I'm sorry, but I'm so lost already. Could somebody please explain it to me? What is u?
Let me put it in a simpler form, just replace u with x something you are probably more common to seeing.

Now the derivative for the ln x = 1/x.

now if you have \int {1/x}dx you know that's the derivative of the ln of x, so you end up with that = ln|x| + C

this is just based of knowing the derivative and antiderivative of ln x, thats all you need to know.

[kreil]

integrals at infinity are calculated by

\int_1^{\infty}f(x)dx=\lim_{t{\rightarrow}\infty}\ int_1^tf(x)dx

if you use this in combination with the info above you can calculate it

[JasonRox]

integrals at infinity are calculated by

\int_1^{\infty}f(x)dx=\lim_{t{\rightarrow}\infty}\ int_1^tf(x)dx

if you use this in combination with the info above you can calculate it

I never learned that yet. That's pretty cool.

[Aki]

so basically, there's not "number" answer to that questions? The answer is just a function?
and also where did ln(x) come from?

[vincentchan]

\int_1^{\infty} \frac{1}{x}dx is undefined, or infinite.. depend on which one you feel more comfortable

where did ln x came from...hmmm... it came from [itex] \frac{d}{dx} lnx = 1/x [/tex].... so your next question is why this is true.....

assume you know product rule and the derivative of e^x is e^x itself

e^{\ln{x}} = x

\frac{d}{dx} e^{\ln{x}} = \frac{d}{dx} x

e^{\ln{x}} \frac{d}{dx} (\ln{x}) =1 --------product rule

x \frac{d}{dx} (\ln{x})=1

\frac{d}{dx} (\ln{x}) = \frac{1}{x}

so the anti-derivative of 1/x is ln(x)

[Zurtex]

Erm vincentchan don't you mean the chain rule?

[dextercioby]

Yes,it is the chain rule...Anyway,the result is correct and the method of finding it is correct as well...

Daniel.