Comparison Theorem for integrals

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SUMMARY

The discussion focuses on using the Comparison Theorem to determine the convergence or divergence of the integral of the function f(x) = sqrt(1 + sqrt(x))/sqrt(x) dx. The key approach involves selecting a comparison function g(x) such that the limit lim (f/g) as x approaches infinity equals a positive constant k. The leading terms of f(x) simplify to f(x) ~ x^(-1/4), suggesting g(x) = x^(-1/4) as an effective choice for the limit comparison test. The conclusion confirms that since the integral of g(x) diverges, the original function also diverges.

PREREQUISITES
  • Understanding of the Comparison Theorem in calculus
  • Familiarity with limit comparison tests
  • Knowledge of integral calculus, specifically improper integrals
  • Ability to manipulate functions in radical form
NEXT STEPS
  • Study the Comparison Theorem for integrals in detail
  • Learn about limit comparison tests and their applications
  • Explore techniques for integrating functions in radical form
  • Investigate the convergence and divergence of improper integrals
USEFUL FOR

Students and educators in calculus, mathematicians focusing on integral convergence, and anyone seeking to deepen their understanding of the Comparison Theorem and its applications in integral calculus.

Corky
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I have to use the comparison theorem to find if the function converges or diverges. Any ideas as to what I can use to compare the function to??

integrate: sqrt(1 + sqrt(x))/sqrt(x)dx
Using the comparison theorem!
 
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You need a function g(x) for which your function (call it f(x)) satisfies the following:

limx-->∞(f/g)=k

where k is a positive real number.

Look at the leading terms in the numerator and denominator of f(x). Rewriting your function f(x) in radical form, it looks like.

f(x)=(x1/2+1)/x1/2

For large x, we can ignore the "+1" in the numerator to get:

f(x)~x1/4/x1/2=x-1/4

g(x) looks to me like a good choice for your test function. Why? Because it is easy to establish the convergence of that integral (I assume you are integrating from 1 to infinity or something like that). Try the limit comparison test on that and see what happens.

Also, your original function is easy to integrate, so you can check your answer. Just let u=x1/2 and so du=(1/2)x-1/2.

edit: fixed HTML code
 
(1 + x^1/2)/x <= ((1 + x^1/2)/x)^1/2 for all x >= 1.
Thus since the integral of(1 + x^1/2)/x diverges as does the original function.
Thanks
 

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