Help: Proving the tending of an integral

  • Thread starter Thread starter Beowulf2007
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The integral f(t) = ∫t2t e-x² dx approaches zero as t tends to infinity. The proof utilizes the Integral Mean Value Theorem, which states that for a continuous function f(x) on [a, b], there exists a number c in (a, b) such that ∫ab f(t) dt = f(c)(b-a). By applying this theorem, it is shown that f(t) = te-c², where c lies between t and 2t, leading to the conclusion that the limit of te-t² as t approaches infinity is zero.

PREREQUISITES
  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with the Integral Mean Value Theorem.
  • Knowledge of limits and asymptotic behavior of functions.
  • Basic proficiency in exponential functions and their properties.
NEXT STEPS
  • Study the Integral Mean Value Theorem in detail.
  • Explore the behavior of the function e-x² as x approaches infinity.
  • Learn about improper integrals and their convergence criteria.
  • Investigate other applications of limits in calculus, particularly in relation to integrals.
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and limit proofs, as well as educators looking for examples of applying the Integral Mean Value Theorem in proofs.

Beowulf2007
Messages
15
Reaction score
0

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?

Homework Equations





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.
 
Physics news on Phys.org
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?

Homework Equations





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,
Beowulf and thanks your reply

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:

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
Replies
34
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
6
Views
3K