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- anonymous

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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- anonymous

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- anonymous

You have to evaluate the limits.

- anonymous

yeah...i evaluate it like any other integrals then plugin the limits right?

- anonymous

Yes, they are solved like normal integrals. assuming the functions are continuous.

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- anonymous

Comparison 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 sub-case of the infinite interval integral, however versions of the test exist for the other sub-cases of the infinite interval integrals as well as integrals with discontinuous integrands.
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

- anonymous

@medisynergi how do i know if it is continuous..and @Rohangrr what? you lost me :p

- anonymous

By definition. for example | x | has to be evaluated separately on the inerval -\inf to \inf

- anonymous

1/ (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.

- anonymous

@Rohangrr: Why do you think that is relevant here?

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