Calculus...

[RoyalXBlood]

The problem is the integral of 1/[math]x^{.2}[/math] from x=0 to x=8. Apparently it's not divergent, but the problem 1/[math]x^{1.4}[/math] is and I understand that. Why is one divergent and not the other, and how would I get the area of 1/[math]x^{.2}[/math] from x=0 to x=8?

[mooeypoo]

Hi,

 

First, I think it might help you to review what integration means, in terms of the geometry of graphs. I don't quite want to just give you an answer what the area under the curve is, because I don't think that will help you solve other questions like it, so here's a good reference on Area under the curve: http://www.intmath.com/Applications-integration/2_Area-under-curve.php it also includes a description of where this formula came from, so it can help you understand the principles.

 

As for divergence, I learned that with relation to series, but integrals and series are intertwined, in a sense, so here's another resource about divergence and convergence of improper integrals: http://www.sosmath.com/calculus/improper/convdiv/convdiv.html

 

What helped me when I studied convergence and divergence, was visualizing the limits.

If an improper integral has a limit, it is convergent.

If an improper integral has no limit, or a limit of infinity, it is divergent.

 

If you think about it, a limit that has a value means that the "graph" of that integral is slowly getting closer and closer to a value of a number -- it converges on a value. If the integral has no limit or limit of infinity, it doesn't have a specific place it gets 'close' to, therefore it is divergent - it moves away into a nonspecific point in space.

 

It's a bit of a simplification but that helped me visualizing it.

 

These links are good for the basics of both subjects.. I just am not sure where, exactly, to start with an explanation. So read the resources, they should - at the very least - help you put things to order in your mind.

 

If you still have questions, feel free to ask

 

~moo

[RoyalXBlood]

But isn't the limit of 1/[math]x^{.2}[/math] as x goes to zero equal to infinite?

[Josy]

Yes it is, but I think you were probably asking about the integral of [math]\frac{1}{x^2}[/math] which also goes to infinity. If you write it with a negative index it becomes much clearer. It converges if you let the upper limit go to infinity though. Have you read the question correctly? I've been caught out by that sort of mistake more often than I care to recall.