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 2 years ago
uhmmm so let me get this straight...improper integrals are solved like normal integrals then plug in values of infinity and the thingies??? @lalaly
 2 years ago
uhmmm so let me get this straight...improper integrals are solved like normal integrals then plug in values of infinity and the thingies??? @lalaly

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medisynergi
 2 years ago
Best ResponseYou've already chosen the best response.2You have to evaluate the limits.

ladiesman123
 2 years ago
Best ResponseYou've already chosen the best response.0yeah...i evaluate it like any other integrals then plugin the limits right?

medisynergi
 2 years ago
Best ResponseYou've already chosen the best response.2Yes, they are solved like normal integrals. assuming the functions are continuous.

Rohangrr
 2 years ago
Best ResponseYou've already chosen the best response.2Comparison Test for Improper Integrals Now that we’ve seen how to actually compute improper integrals we need to address one more topic about them. Often we aren’t concerned with the actual value of these integrals. Instead we might only be interested in whether the integral is convergent or divergent. Also, there will be some integrals that we simply won’t be able to integrate and yet we would still like to know if they converge or diverge. To deal with this we’ve got a test for convergence or divergence that we can use to help us answer the question of convergence for an improper integral. We will give this test only for a subcase of the infinite interval integral, however versions of the test exist for the other subcases of the infinite interval integrals as well as integrals with discontinuous integrands. http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

ladiesman123
 2 years ago
Best ResponseYou've already chosen the best response.0@medisynergi how do i know if it is continuous..and @Rohangrr what? you lost me :p

medisynergi
 2 years ago
Best ResponseYou've already chosen the best response.2By definition. for example  x  has to be evaluated separately on the inerval \inf to \inf

medisynergi
 2 years ago
Best ResponseYou've already chosen the best response.21/ (1  x ) is not defined at x = 1. It is not continuous at that point. so the interval 0 to \inf would make no sense.

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.0@Rohangrr: Why do you think that is relevant here?
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