How large must a be for the improper integral to be less than .001?

In summary, the conversation discusses how to determine the value of "a" in order for the integral of 1/(x^2+1) from "a" to infinity to be less than 0.001. The method involves using the potential equation and taking the limit of the equation as t goes to infinity. The final answer is pi/2.
  • #1
Omega_Prime
23
0
Been doing some calculus review to knock the rust off for this coming fall semester and I got stuck...

Homework Statement



From Stewart's book (Early Transcendentals: 6E): (7.8 pg517 #69)

Determine how large the number "a" has to be so that:

[itex]\int[/itex][itex]\stackrel{\infty}{a}[/itex][itex]\frac{1}{x^{2}+1}[/itex]dx <.001

Homework Equations



None.

The Attempt at a Solution



Ok, I can easily picture the graph and the area under it. I figured I'd integrate, use "a" for my lower bound and "t" for the upper bound, then by using the potential equation it's just a simple matter of solving for "a" while taking the limit of said equation as t goes to infinity.

I managed to get:

arctan (t) - arctan (a) < 1/1000 (I suck at "latex" but this should technically be the limit of those arctans as t -> infinity < .001)

Here is where I think I'm screwing up... I take the tangent of both sides:

tan [arctan t -arctan a] < tan (1/1000) - and I'm stuck, I know I can't just apply the tangent function independently to both parameters giving me t - a < tan (.001) is there some trig identity I'm not thinking of..?
 
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  • #2
Take the limit as t → ∞ .

[tex]\lim_{t\to\infty}\,\arctan(t) = ?[/tex]
 
  • #3
Ugh... Sorry to waste your time. The limit is pi/2, all I had to do was take the limit and I was already done lol. For some reason I thought there would be no limit for the arctan and that was bothering me.
 

1. What is an improper integral?

An improper integral is an integral where one or both of the bounds of integration are infinite or the integrand is undefined at some point within the bounds of integration.

2. How do you evaluate improper integrals?

Improper integrals can be evaluated by breaking them into smaller integrals and taking the limit as the bounds of integration approach infinity or the undefined point. Alternatively, you can use techniques such as the comparison test or the limit comparison test to determine if the integral converges or diverges.

3. What is the difference between a convergent and divergent improper integral?

A convergent improper integral is one where the limit of the integral exists and is finite, while a divergent improper integral is one where the limit does not exist or is infinite.

4. Can improper integrals be solved using the Fundamental Theorem of Calculus?

Yes, the Fundamental Theorem of Calculus can still be applied to improper integrals as long as the integrand is continuous on the interval of integration.

5. What are some real-world applications of improper integrals?

Improper integrals are used in many fields, including physics, engineering, and economics, to model and solve problems involving infinite or undefined quantities. Examples include calculating the total energy of a system with an infinite number of particles or determining the value of a continuous income stream that goes on indefinitely.

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