Comparisons of imporper integrals

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In summary, the conversation is discussing finding the upper bound for the integral of e^(-x^2) from 3 to infinity. The hint given is to use the fact that e^(-x^2) is smaller than e^(-3x) for x>3. After some calculations, the upper bound is determined to be 1/3 e^(-9).
  • #1
ada0713
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Homework Statement


Find the upper bound for

[tex]\int[/tex] (from 3 to infinity) e^(-x^2)

[hint: e^(-x^2) < e^(-3x) for x>3]


The Attempt at a Solution



I don't understand what they mean by "upper bound"
Are they asking for the upper bound for the area below e^(-x^2)
for 3<x<infinity ?

I knew that the function converges by graphing the function
y= e^(-x^2) and y=e^(-x) and comparing between those two,
but I'm not sure what the problem is asking for.
Pleas help me witht the start!
 
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  • #2
If for two functions [itex]f,\,g[/itex] you have [itex]f(x)<g(x),\, \forall x>x_0[/itex] what can you say about the integrals [itex]\int_{x_0}^\alpha f(x)\,d\,x,\,\int_{x_0}^\alpha g(x)\,d\,x[/itex]?
 
  • #3
That [itex]\int_{x_0}^\alpha f(x)\,d\,x,\ < \int_{x_0}^\alpha g(x)\,d\,x[/itex] and that if
\int_{x_0}^\alpha g(x)\,d\,x[/itex] converges, then [itex]\int_{x_0}^\alpha f(x)\,d\,x,\ converges as well?

But how do I find upper bound knowing this?
 
  • #4
I meant "integral of g(x)" for the first one and "integral for f(X)" for the second one. Sorry about the mess
 
  • #5
Let [itex]f(x)=e^{-x^2},\,g(x)=e^{-3\,x}[/itex]. Can you calculate
[tex]\int_3^\infty g(x)\,d\,x[/tex]?
 
  • #6
Yes, I calculaed it and
lim[b to infinity] (-1/3 e^(-5b)) (1/3)
so the first part goes to zero when you plug in infinity for b
therefore it convererges to 1/3

.. so.. is it right that since f(x) is smaller than g(x) it converges as well
and therefore the possible upper bound should be 1/3 (or smaller than 1/3) ?
 
  • #7
I think the result is
[tex]\int_3^\infty e^{-3\,x}\,d\,x=\frac{1}{3}\,e^{-9}[/tex] :smile:
And the answer is that one upper bound is [itex]\frac{1}{3}\,e^{-9}[/itex], i.e.

[tex]\int_3^\infty e^{-x^2}\,d\,x<\frac{1}{3}\,e^{-9}[/tex]
 
Last edited:
  • #8
ooops,, I guess I made a calculation mistake somewhere in the middle, I'll figure it out what I did wrong. and THANK YOU for your help!
 
  • #9
Your are faster in typing! :smile:
The result I posted it's the correct one, and one upper bound is [itex]\frac{1}{3}\,e^{-9}[/itex], or every other number bigger than that, not smaller!

Glad I helped :smile:
 
  • #10
I noticed that you meant 1/3 e^(-9)
:) Thanks again!
 

Related to Comparisons of imporper integrals

1. What is an improper integral?

An improper integral is a type of integral where either the upper or lower limit of integration is infinite, or the function being integrated is unbounded at one or more points in the interval of integration. This means that the area under the curve cannot be calculated using the traditional methods of integration and requires special techniques.

2. How is an improper integral different from a regular integral?

A regular integral has both the upper and lower limits of integration within a finite interval and the function being integrated is also bounded within that interval. This allows for the use of traditional integration methods such as the Fundamental Theorem of Calculus. Improper integrals, on the other hand, have one or both limits of integration at infinity or have an unbounded function, which requires different methods of calculation.

3. What are some common examples of improper integrals?

Some common examples of improper integrals include integrals with infinite limits such as ∫1 1/x dx, integrals with unbounded functions such as ∫01 1/√x dx, and integrals with discontinuous functions such as ∫01 1/x dx.

4. How do you evaluate an improper integral?

To evaluate an improper integral, you can use one of several methods such as the limit definition of an integral, comparison test, or the Cauchy principal value. These methods involve taking the limit of the integral as the limit of integration approaches infinity or a point where the function is unbounded.

5. Why are improper integrals important in science?

Improper integrals are important in science because many real-world problems involve infinite quantities or unbounded functions. For example, in physics, improper integrals are used to calculate the work done by a variable force or the total energy of a system. In economics, they are used to model infinite demand or supply curves. Improper integrals also play a crucial role in probability and statistics, particularly in calculating expected values of continuous random variables.

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