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

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SUMMARY

The discussion focuses on determining the value of "a" such that the improper integral \(\int_{a}^{\infty} \frac{1}{x^{2}+1}dx < 0.001\). The user successfully derived the equation \(\arctan(t) - \arctan(a) < 0.001\) and realized that the limit of \(\arctan(t)\) as \(t\) approaches infinity is \(\frac{\pi}{2}\). This insight allowed the user to conclude that they could solve for "a" directly, eliminating previous confusion regarding the tangent function and its application.

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  • Understanding of improper integrals
  • Familiarity with the arctangent function and its limits
  • Basic knowledge of calculus, specifically integration techniques
  • Ability to manipulate inequalities involving trigonometric functions
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Omega_Prime
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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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Take the limit as t → ∞ .

[tex]\lim_{t\to\infty}\,\arctan(t) = ?[/tex]
 
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.
 

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