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Improper integrals and comparison tests

I have this problem:  Evaluate the integral (x^2-2)/(x^4+3) from 1 to infinity using a comparison test (evaluate if converging).  I understand the basic idea is to find a function with a graph "above/below" this function that can be integrated.  My solutions guide uses (3x^2)/(x^4) which of course is just 3x^-2.  I see that it's easy to find the antiderivative of this and take it's limit at infinity.  How did they come up with 3x^-2 as a comparison?  Am I just missing some simple Algebra?

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Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (275 lesson ratings) (275)

Notice that the numerator is positive for all x > √2 and the denominator is positive for all real x.

In a fraction with both numerator and denominator being positive, increasing the numerator or decreasing the denominator will increase the value.

e.g. 2/3 > 2/5 because 3 < 5 and 3/5 > 2/5 because 3 > 2.

So using this when x > √2:

We can increase the numerator by 2 to x2, and decrease the denominator by 3 to x4.

We wind up with x2/ x4 > (x2-2) / (x4+3).

So we can use x2/x4 = x-2 as an upper bound for this interval. (3x-2 is usable too since it is even bigger than x-2).

The integrand is negative for 1<x<√2 so our bound applies in that region too and therefore is an upper bound for all x>1.

Thus ∫1(x2-2)/(x4+3) dx < ∫1 x-2 dx = [-x-1]1 = 0 - (-1) = 1, which proves convergence.



OK...  I think I got it.  So as long as the comparison function value (in this case) is greater than the original function in the bounds of the integral, the comparison will work for convergence/divergence...  I can manipulate the fraction as necessary for ease of finding the antiderivative provided this is kept this true.

Yes. And if the integral diverges, you would look for a lower bound that can be easily shown to diverge. This assumes that the integrand is positive valued in the interval where convergence is unclear. For example, in the portion of the integration interval where you integrand is negative 1<x<v2, the integral converges, so it suffices to show that it converges on interval x>v2 where your integrand is positive.