Takeover Season
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Hi,

So, I know how to do (ii), but it's (iii) that usually gets the best of me.
SORRY, IT DOESN'T SHOW FOR SOME REASON: BUT, (ii) is calculating F'(x) and (iii) is the last line, i.e. using that to find F(x).
Once we have found F'(x), what are the possible ways of finding F(x)?

I am thinking: One way is to use the FTC and just say \int _{x=a}^X\:F'\left(x\right)dx\:=\  :F\left(X\right)-F\left(a\right) and if we choose x = a such that F(a) = 0, we are done as we have calculated F'(x) from (ii) already.

My question is: How do we find that x = a, here it is quite easy, x = 1 makes the integrand and hence the entire integral F(x) = 0. But, what if we couldn't find an x, so easily, that makes it = 0, how would we approach it then instead?

Also, are there any alternative methods anyone can think of? (Not in this question as it specifically asks to find F'(x) and proceed, but in general?

Thank you
Last edited by Takeover Season; 1 month ago
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RDKGames
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(Original post by Takeover Season)
Name:  Screen Shot 2020-08-05 at 17.04.26.png
Views: 7
Size:  164.5 KB

Hi,

So, I know how to do (ii), but it's (iii) that usually gets the best of me.
SORRY, IT DOESN'T SHOW FOR SOME REASON: BUT, (ii) is calculating F'(x) and (iii) is the last line, i.e. using that to find F(x).
Once we have found F'(x), what are the possible ways of finding F(x)?

I am thinking: One way is to use the FTC and just say \int _{x=a}^X\:F'\left(x\right)dx\:=\  :F\left(X\right)-F\left(a\right) and if we choose x = a such that F(a) = 0, we are done as we have calculated F'(x) from (ii) already.

My question is: How do we find that x = a, here it is quite easy, x = 1 makes the integrand and hence the entire integral F(x) = 0. But, what if we couldn't find an x, so easily, that makes it = 0, how would we approach it then instead?

Also, are there any alternative methods anyone can think of? (Not in this question as it specifically asks to find F'(x) and proceed, but in general?

Thank you
This is indeed a fine approach.

In the case where no such value of a is obvious, you wouldn't be able to be more specific than that.

In some cases you might be able to show the existance of a special value a=\alpha such that F(\alpha) = 0 but not know what it is exactly, in which case you can still use it as it is.
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Takeover Season
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(Original post by RDKGames)
This is indeed a fine approach.

In the case where no such value of a is obvious, you wouldn't be able to be more specific than that.

In some cases you might be able to show the existance of a special value a=\alpha such that F(\alpha) = 0 but not know what it is exactly, in which case you can still use it as it is.
Oh, thank you. So, I guess, there 'has' to be an a for which F(a) = 0 for this method to work?
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