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anonymous
 5 years ago
Suppose f is continuous on [0, infinity) and limit of f(x) as x approaches infinity is 1. Is it possible that the integral of f(x) from 0 to infinity is convergent?
anonymous
 5 years ago
Suppose f is continuous on [0, infinity) and limit of f(x) as x approaches infinity is 1. Is it possible that the integral of f(x) from 0 to infinity is convergent?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{a \rightarrow \infty} \int\limits_{0}^{a} f(x)dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you show me the steps to proof that it is possible the integral is convergent? cos right now i only have the answer that it is not convergent by using integration by parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Depending that F(x) integrates into a simple equation in terms of x, then yes it would converge because you would take x, replace it with a, and then use your limit and take it to infinity. Easy. It would diverge if the infinity would be in the denominator of a quotient and the denominator was going to infinity faster than the numerator. Just look at the big picture, or "in a long run".

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 1/x^p and p > 1, then it CONVERGES.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi quantish, if you take f(x) as 1/x^p and p>1, what does it converge to? I get a divergent answer when i integrate that from 0 to infinity. Only when I integrate it from 1 to infinity do i get a convergent answer. thanks so much for all the help everyone.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was just refering to the pseries test, and wasn't refering to a function in particular and only a limit as x > infinity. If the denominator increases at a greater rate than the numerator i.e. 1/x^p when p > 1, then it should CONVERGE, not DIVERGE as was earlier stated.
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