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    Really struggling with part b of this question, not sure how to get b involved or get anywhere other than where I am. Any pointers are appreciated!

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    (Original post by Substitution)
    Really struggling with part b of this question, not sure how to get b involved or get anywhere other than where I am. Any pointers are appreciated!

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    What does your partition for part a) look like?
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    (Original post by SeanFM)
    What does your partition for part a) look like?
    My set is Pn={xn=0,x1,x2,...,xn-1,xn=b}

    Thanks for the reply
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    (Original post by Substitution)
    My set is Pn={xn=0,x1,x2,...,xn-1,xn=b}

    Thanks for the reply
    You're correct in that x0 = 0 and xn = b.

    Otherwise, what is the expression for the nth term? (i.e what is x1, x2 etc).
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    (Original post by SeanFM)
    You're correct in that x0 = 0 and xn = b.

    Otherwise, what is the expression for the nth term? (i.e what is x1, x2 etc).
    Is the nth term just (b/n)*(the partition number x1,x2 etc)?
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    (Original post by Substitution)
    Is the nth term just (b/n)*(the partition number x1,x2 etc)?
    Correct, yes. Kind of annoying that they've defined n as the number of intervals rather than 'N', so it becomes a bit confusing if you define the points as xn, though I suppose you could use xN = (b/n)*N.

    Does that show you how to do part b?
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    (Original post by SeanFM)
    Correct, yes. Kind of annoying that they've defined n as the number of intervals rather than 'N', so it becomes a bit confusing if you define the points as xn, though I suppose you could use xN = (b/n)*N.

    Does that show you how to do part b?
    Still think I must be going wrong someone or not seeing something, my sun has no relation to the formula in question. Thanks for your help

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    (Original post by Substitution)
    Still think I must be going wrong someone or not seeing something, my sun has no relation to the formula in question. Thanks for your help

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    You are getting there just don't lose sight of the original question, that you're looking for the lower and upper Riemann sums.

    For both of those things, it won't be the sum to b terms, but something else. Can you think of what it should be?

    And you can take any constants out of sums as well, so remember what's constant and what's not. (Eg the sum of n lots of 3 is the same as 3 * n lots of 1.

    The sum you've got on that page looks almost like the upper Riemann sum, with the problem being that it's the sum to b terms rather than the sum to ... terms.

    And then use the rest of what I've said in this post.
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    (Original post by SeanFM)
    You are getting there just don't lose sight of the original question, that you're looking for the lower and upper Riemann sums.

    For both of those things, it won't be the sum to b terms, but something else. Can you think of what it should be?

    And you can take any constants out of sums as well, so remember what's constant and what's not. (Eg the sum of n lots of 3 is the same as 3 * n lots of 1.

    The sum you've got on that page looks almost like the upper Riemann sum, with the problem being that it's the sum to b terms rather than the sum to ... terms.

    And then use the rest of what I've said in this post.
    Ahh that was really helpful thank you! I've got the upper sum and finally understand it!

    The lower sum on the other hand is a bit more difficult and I can't figure out what to change to calculate it, would it be the sum of n-1 terms? Thanks
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    (Original post by Substitution)
    Ahh that was really helpful thank you! I've got the upper sum and finally understand it!

    The lower sum on the other hand is a bit more difficult and I can't figure out what to change to calculate it, would it be the sum of n-1 terms? Thanks
    Brilliant :borat:

    I think that you can manipulate it into something that is the sum of n-1 terms of something, yes.
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    I wonder what happened to DFranklin
 
 
 
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