Does anyone have a nice example of an improper integral over a vertical asymptote that does result in a value, suitable for FM students? (Something like the integral from -1 to 1 of 1/x^2, but that's not undefined). I'd rather avoid piecewise defined or modulus functions if possible. #ALevelMaths - ThreadSky

susanwhitehouse.bsky.social • 196 days ago

Does anyone have a nice example of an improper integral over a vertical asymptote that does result in a value, suitable for FM students? (Something like the integral from -1 to 1 of 1/x^2, but that's not undefined). I'd rather avoid piecewise defined or modulus functions if possible. #ALevelMaths

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rufuswilliam.bsky.social•195 days ago

Hey Susan, can I ask you some questions that might make me sound like I'm stupid? I've bought your book.

susanwhitehouse.bsky.social•195 days ago

Ask any questions you like 😀

rufuswilliam.bsky.social•195 days ago

Thanks! N choose r... To people generally understand how this works? I don't! I just use it. Can you explain it simply?

susanwhitehouse.bsky.social•195 days ago

I think I'm going to have to use numbers in the first instance, and then generalise to n and r. Supposing we want to choose 2 things from 6 different things. There are 6 ways of choosing the first one, and then 5 ways of choosing the second. That gives us 6 x 5 which can be written as 6!/4!

rufuswilliam.bsky.social•195 days ago

Got that

susanwhitehouse.bsky.social•195 days ago

But then we have repeats because we have counted choosing A then B as well as choosing B then A. So we need to divide that answer by 2!, giving us the result of 6!/4!2!. Now let's think about choosing r objects from n. For the first object we have n possibilities, then n-1, then n-2 etc.

rufuswilliam.bsky.social•195 days ago

Bang bang..it's easy..thank you!

susanwhitehouse.bsky.social•195 days ago

This can be written as n!/(n-r)!. But we will have those pesky repeats, so we need to divide by r! to get rid of them, leading to n!/r!(n-r)!

mrconnormaths.bsky.social•195 days ago

📌

over-drawn.bsky.social•196 days ago

It's a little straight forward perhaps but 1/x^(1/3) converges between, say, x=-2 and x=3.

susanwhitehouse.bsky.social•196 days ago

Yes, that works, thanks.

paddymacmahon.com•196 days ago

I'll update these questions to include something that goes across an asymptote too, I think. Thanks for the suggestion!
https://www.paddymacmahon.com/resources/a-level/pure#h.pqhct7yw84h9

paddymacmahon.com•196 days ago

Something like this... unfortunately the right-hand intersection isn't terribly nice!
#ALevelMaths

paddymacmahon.com•196 days ago

Nicer limits and an easier question!

susanwhitehouse.bsky.social•196 days ago

I had just kind of forgotten that cube roots, fifth roots etc, are defined for negative numbers as well as for positive numbers 🤦 I had thought that I was stuck because I couldn't use a power of x between -1 and 0 on both sides of the asymptote. Basically, I'm an idiot.

liemers.bsky.social•195 days ago

📌

mathforge.org•195 days ago

In your defence, they *shouldn't* be definable for negatives - it's one of nature's exceptions that they happen to work but it's not something that should be talked about in polite company.

susanwhitehouse.bsky.social•195 days ago

I appreciate you coming to my defence!

paddymacmahon.com•195 days ago

Agreed!

susanwhitehouse.bsky.social•196 days ago

Thanks 😀

drmthornber.bsky.social•196 days ago

Try x^(-2/5) from -1 to 1. Both infinitely tall parts have positive area.

susanwhitehouse.bsky.social•196 days ago

Basically I need to integrate x^(- even number/bigger odd number) over the asymptote x=0, and I should be alright, shouldn't I?

susanwhitehouse.bsky.social•196 days ago

Yes, that's nice, thanks 😀

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