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Finding upper and lower sums of a region

  1. Dec 6, 2004 #1
    here is my problem: find the upper and lower sums for the region bounded by the graph of [tex]f(x) = x^2[/tex] and the x-axis between x=0 and x=2. I understand what this problem is asking but i don't understand how to compte the left and right endpoints. the left endpoint is the following:

    [tex]m_{i}=0+(i-1)\frac{2}_{n}[/tex] = [tex]\frac{2(i-1)}_{n}[/tex]

    and the right enpoint is given as this:

    [tex]M_{i}=0+i\frac{2}_{n}[/tex] = [tex]\frac{2i}_{n}[/tex]

    The lower sum ( left enpoint) is the following:

    8\3 - 4\n + 4\3n^2

    and the right endpoint is computed in a similar fashion.

    my question is, how do i find the left and right enpoints? what is the formula for doing this? what if the lower bound is not zero but some other number, how would i find it then?
     
  2. jcsd
  3. Dec 7, 2004 #2

    shmoe

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    The lower (and upper) endpoints depend on n. For a given n, the [itex]m_i[/itex] you gave gives all the left endpoints if you let i range from 1 to n (same for the upper sum). Maybe working out a couple examples wouldn't hurt, if n=2 or 3 can you find the endpoints?

    What are you planning on using for the lower sum?

    In your formulas for the left and right endpoints, the 0 is the leftmost point in the interval, the "2/n" is signifying the width of each sub-interval. If you are splitting the interval [a,b] up into n pieces, what would you replace these with?
     
  4. Dec 7, 2004 #3

    HallsofIvy

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    I don't understand your question: you say, correctly, that
    "the left endpoint is the following:

    = [tex]\frac{2(i-1)}_{n}[/tex]"
    and
    "the right endpoint is given as this
    [tex]\frac{2i}_{n}[/tex]"

    So what could you possibly mean by "how do i find the left and right enpoints?" Didn't you just do that?
     
  5. Dec 7, 2004 #4
    this problem solution was copied from my text. i did not fing the endpoints myself.
     
  6. Dec 7, 2004 #5

    HallsofIvy

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    You are asked to find the integral from 0 to 2 and are dividing that into n intervals.
    The length of each interval, therefore, is the distance from 0 to 2 (which is 2) divided by n: 2/n. For example, if you only used 2 intervals, each would have length 1, of course. And your "endpoints" would be 0, 1, 2: start from the left at 0 and add that length. If you used 4 intervals, each would have length 2/4= 1/2 and your endpoints would be 0+ 0*(1/2)= 0, 0+ 1*(1/2)= 1/2, 0+ 2*(1/2)= 1, 0+ 3*(1/2)= 3/2, and 0+ 4*(1/2)= 2. If you use n intervals, each would have length 2/n and your endpoints would be 0+ 0*(2/n)=0, 0+ 1*(2/n)= 2/n, ..., 0+ i*(1/n)= 2/n.

    To get the right endpoints start with i= 1 and go up to n. To get the left endpoints, you could just use i= 0 to n-1. What your text is doing apparently, is replacing i in
    2i/n by i-1 which gives 2(i-1)/n and you can use i= 1 to n again.
     
  7. Dec 8, 2004 #6
    thanks IVY, I still don't quite understand what's going on with this idea,but i really appreciate the feedback.
     
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