Does this integral exist?

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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.
 

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  • #2
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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  • #3
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?
 
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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.
 
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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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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.
 
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  • #7
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##.
 
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  • #8
Samy_A
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I think that @Vsmith196 's teacher expected the examinees to just assume that ##f(0)=0##, without giving it a second thought. Kudos to Vsmith196 for noticing that this choice, while "reasonable", is still a choice that has to be made.
 
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