# Does this integral exist?

I was asked to take the integral of (x^4 -x^6) / (x^2) from 0 to 4. Does this exist? Although there is a hole at 0, the limit exists. And the denominator factors out. It's not an asymptote. Please let me know. I can see both reasons why it would and why it wouldn't.

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.

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• Vsmith196
PeroK
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I was asked to take the integral of (x^4 -x^6) / (x^2) from 0 to 4. Does this exist? Although there is a hole at 0, the limit exists. And the denominator factors out. It's not an asymptote. Please let me know. I can see both reasons why it would and why it wouldn't.

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?

• Vsmith196
I'm not in calculus II yet. I haven't learned improper integrals.
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.

You should use the definition of an improper integral

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

PeroK
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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.

• geoffrey159
Samy_A
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All I've been taught is that if the problem isn't continuous due to a vertical asymptote, it doesnt exist.
On a side note, a definite integral can exist even with a vertical asymptote.
For example ##\int_{0}^{1}\frac {1}{\sqrt{x}}dx=2##, and the function has a vertical asymptote at ##x=0##.

• geoffrey159
Samy_A
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