# Help: Proving the tending of an integral

1. Mar 3, 2008

### Beowulf2007

1. The problem statement, all variables and given/known data

Given the integral $$f(t) = \int_{t}^{2t} e^{-x^2} dx$$

How do I then prove that f(t) = 0 if t tends to infinity?

2. Relevant equations

3. The attempt at a solution

I can that if I make t goes from minus infinity to zero, then the limit will tend to 1, but when making it tend to infinity then I get a limit which is zero.

But this cannot be proof enough can it?

Cheers,
Beowulf.

2. Mar 3, 2008

### HallsofIvy

Staff Emeritus
Well, it might be! How do you "get a limit which is zero"?

I would be inclined to use the "integral mean value theorem":
"If f(x) is continuous on [a, b], then there is at least one number c in (a, b) for which
$$\int_a^b f(t)dt= f(c)(b-a)$$".

So
$$\int_t^{2t} e^{-x^2} dt= e^{-c^2}(2t- t)= te^{-c^2}$$
where c lies between t and 2t. The crucial point is that c> t so that, for t> 0,
$$0< te^{-c^2}< te^{-t^2}$$
Now, what is the limit of that as t goes to infinity?

3. Mar 3, 2008

### Beowulf2007

Hi

Then limit must be

$$\mathop{\lim} \limits_{t \to \infty} te^{-t^2} = 0$$, then assuming this is true then f(x1) an f(x2) then tend to the same limit since f is continious (by the mvt), and therefore
the integral f(t) -> 0, then t tends to infinity?

q.e.d.

Cheers,