# Help: Proving the tending of an integral

• Beowulf2007
In summary, the conversation discusses how to prove that the integral f(t) = \int_{t}^{2t} e^{-x^2} dx equals 0 as t tends to infinity. The discussion involves using the "integral mean value theorem" and showing that the limit of te^{-t^2} tends to 0 as t goes to infinity. It is concluded that f(t) also tends to 0 as t goes to infinity, assuming that f is continuous by the mvt.
Beowulf2007

## Homework Statement

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?

## 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.

Beowulf2007 said:

## Homework Statement

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?

## 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.
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?

HallsofIvy said:
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?

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,

p.s. How does this look?

p.p.s. Is enough to assume that f is continious by the mvt or do I need to show it part of the proof?

Last edited:

## 1. What is an integral?

An integral is a mathematical concept that represents the accumulation of a quantity over a certain interval. It is often used to find the area under a curve or the total distance traveled by an object.

## 2. How do you prove the tending of an integral?

The tending of an integral can be proven by using the Fundamental Theorem of Calculus, which states that if a function is continuous over a certain interval, then the integral of that function can be evaluated using the antiderivative of the function at the endpoints of the interval.

## 3. What is the significance of proving the tending of an integral?

Proving the tending of an integral is important because it allows us to accurately calculate the area under a curve or the total distance traveled by an object. It also helps us understand the behavior of a function over a given interval.

## 4. What are some common techniques used to prove the tending of an integral?

Some common techniques used to prove the tending of an integral include using the definition of the integral, the Fundamental Theorem of Calculus, and properties of integrals such as linearity and change of variables.

## 5. Can the tending of integrals be used in real-world applications?

Yes, the tending of integrals is used in various real-world applications such as calculating the volume of a 3-dimensional object, determining the average value of a function, and finding the center of mass of an object.

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