Solving Indefinite Integral: F(x)= \int\frac{1}{t}dt from x to 2x

[rdgt3000]

Homework Statement

How is this function continuous from 0 to infinity

F(x) = $$\int\frac{1}{t}$$dt from x to 2x

Homework Equations

I am fairly sure that this equation uses the properties of natural logs to solve.
Also an infinite function has a derivative that is equal to 0.

The Attempt at a Solution

F'(x)= 1/t

That is about how far I have got in solving this problem. I really can't figure out why I am having such problems with this. I think it has to do with the fact that the interval of the integral is from x to 2x.

 

[SiddharthM]

Fundamental theorem of calculus. Let f be a real-valued integrable function defined on a closed interval [a, b]. If F is defined for x in [a, b] by

intregral(1/x) from a to x

then F is continuous on [a, b]. If f is continuous at x in [a, b], then F is differentiable at x, and F ′(x) = f(x).

I presume the problem asks to show that F is continuous on (0,infinity) not [0, infinity] - there's a difference, one is true, the other isn't.

 

[HallsofIvy]

I very much doubt that this "has to do with the fact that the interval of the integral is from x to 2x". If you know the anti-derivative, F(t), then the integral is F(2x)- F(x).

I'm more inclined to think that the problem has to do with the fact that you don't know the anti-derivative of 1/t ! If you do, what is it? If you don't, look it up in your textbook!

 

[rdgt3000]

Sorry I've just been off tonight. Under the attempted solution part I put F'(x)= 1/t . What I meant was the antiderivative was log(t). That one is a no brainer. I'm having some trouble explaining why it's continuous. And yes, it is from the interval (0, infinity), sorry about that. Thank you both for your quick and helpful responses by the way.

 

[Kurdt]

What are the properties of the natural logarithm function?

 

[rdgt3000]

The domain for the natural logarithm is the set of all real numbers and is also differentiable. Therefore the function is continuous from (0, infinity) since the log of any positive number results in an answer that exists, the range is from - infinity to + infinity. I think my problem was from thinking that what if x equaled some negative number, but that is ruled out since the function must occur from (0, infinity). So I am pretty sure I just figured this out, and have also realized that this was a fairly easy question!

 

[Gib Z]

The antiderivative of the integrand in the Natural Log. By the Fundamental theorem on calculus, The integral then evaluates to $F(x) = \log_e (2x) - \log_e (x)$.

Can you see a log property that would simplify the RHS to a constant, making F(x) continuous for all real x?

 

[Kurdt]

Gib Z has better alluded to what I was aiming at. I assumed that you'd actually evaluated the integral after the help given previously. It should be very apparent what the answer is now.