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Samy_A

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The integral exists.

The function you have to integrate is equal to ##x^2-x^4##, and that is a nice polynomial.

The "problem" with ##x=0## is not relevant. You can set your function in ##x=0## to anything you want, that won't change the value of the integral from 0 to 4. Of course, it makes sense to define the function in ##x=0## using the same formula as used for the other points, so you get a function that is continous etc ... everywhere.

Said differently, your function doesn't have a singularity in 0, as 0 is a point of continuity of your function.

The function you have to integrate is equal to ##x^2-x^4##, and that is a nice polynomial.

The "problem" with ##x=0## is not relevant. You can set your function in ##x=0## to anything you want, that won't change the value of the integral from 0 to 4. Of course, it makes sense to define the function in ##x=0## using the same formula as used for the other points, so you get a function that is continous etc ... everywhere.

Said differently, your function doesn't have a singularity in 0, as 0 is a point of continuity of your function.

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You should use the definition of an improper integral:

##\int_{0}^{4}f(x)dx = \lim_{\epsilon \rightarrow 0^+}\int_{\epsilon}^{4}f(x)dx##

Then it's clear?

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You should use the definition of an improper integral:

##\int_{0}^{4}f(x)dx = \lim_{\epsilon \rightarrow 0^+}\int_{\epsilon}^{4}f(x)dx##

Then it's clear?

I think so. Although, I'm still in Calc I, and I havent been taught improper integrals. All I've been taught is that if the problem isn't continuous due to a vertical asymptote, it doesnt exist. This problem was on my test yesterday.

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You should use the definition of an improper integral

This is not an improper integral, as it was clearly explained in previous post.

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This is not an improper integral, as it was clearly explained in previous post.

I guess it depends on your definitions. I'd say ##f## is not continuous at ##x = 0## because ##f(0)## is not defined. It's a removable discontinuity. I can't see anything wrong in tackling a removable discontinuity by treating it as an improper integral.

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Samy_A

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On a side note, a definite integral can exist even with a vertical asymptote.All I've been taught is that if the problem isn't continuous due to a vertical asymptote, it doesnt exist.

For example ##\int_{0}^{1}\frac {1}{\sqrt{x}}dx=2##, and the function has a vertical asymptote at ##x=0##.

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Samy_A

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