Existence of Integral for (x^4 -x^6) / (x^2) from 0 to 4

[Vsmith196]

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]

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 continuous etc ... everywhere.

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

 

[PeroK]

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 haven't been taught improper integrals. All I've been taught is that if the problem isn't continuous due to a vertical asymptote, it doesn't exist. This problem was on my test yesterday.

 

[geoffrey159]

You should use the definition of an improper integral

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

 

[PeroK]

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.

 

[Samy_A]

All I've been taught is that if the problem isn't continuous due to a vertical asymptote, it doesn't 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 $x=0$.

 

[Samy_A]

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.