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Are improper integrals Definite or Indefinite?
The AQA books I am working from are trying to describe it all in half a page, as is often the case. My other book throws some more light, but still confusing.
It gives this as the first example..
Find
Determine the limit
I can understand why the limit would not exist due to x^2 non being a number and diverging.
I don't know.. I'm stuck.:/
Can feel it coming together.. Slowly but Surely
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Hi Dave;
http://www.youtube.com/watch?v=85-HNJyuyrU
http://www.youtube.com/watch?v=KvC4Xyay … re=channel
http://www.youtube.com/watch?v=ISXOUJHj … re=related
The discontinuity can be in between the intervals of integration like this: At x = 0 you are dividing by 0!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobby
Thanks for that, Emma has just been sent to her room for 10 mins for being a bad girl so I will watch some now.
Can feel it coming together.. Slowly but Surely
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Hi Dave;
I am sure there is mathematically tons more about that definition but those video cover what you need to know!
Tell emma there was a bad little boy by the name of bobbym. You only need to say, "Do you want to grow up and be like him?" That usually frightens children ( and teens ) into doing the right thing.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobby
Nice one.. I'll give that a try, she has woken up in a funny mood and already been sent to her room once.
I've only watched the first one so far, and then had another go with the book. I think I have it, though I'm going to look at the rest to be sure.
Is it right that you choose
if the asymptote stretches out to
and
when stretching out to
and that the the area under the curve can only be found when the limit converges on 0, otherwise it is divergent and the area is NaN.
I don't know, maybe I am pushing too hard. I only read about these for the first time last thursday, and 10 months ago I couldn't even add two fractions.
Still I prefer to go on, I like it.:)
Can feel it coming together.. Slowly but Surely
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Hi Dave;
No, I would do it like this:
and that the the area under the curve can only be found when the limit converges on 0, otherwise it is divergent and the area is NaN.
The limit could be a number like 13 and that would mean the integral exists. Not just 0.
I don't know, maybe I am pushing too hard. I only read about these for the first time last thursday, and 10 months ago I couldn't even add two fractions.
I don't see any reason why you shouldn't push as hard as you can. I would slow down when you start talking to yourself or leaving messages to yourself. If you begin to hide money from yourself or steal money from yourself then I would take a vacation.
When you get a chance watch the other vids and some of the others on the sides of the page. Then we will talk about them.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobby
Hi Dave;
No, I would do it like this:
Sorry about that, a typo.
Thank you for this.
Can feel it coming together.. Slowly but Surely
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Hi Dave;
No problem!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I just thought of something. I'm sure there is probably a way around it, though I haven't come across it yet.
What happens when you need to integrate
how do you get around the division by 0
Can feel it coming together.. Slowly but Surely
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Hi Dave;
You don't that integral does not exist.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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