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Math Help - Finding the area under a graph using Riemann Sum question

  1. #1
    Newbie br1tt204's Avatar
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    Finding the area under a graph using Riemann Sum question

    I would appreciate any help I can get with this problem! My prof usually gives us a few practice problems we should know. One of them this week was finding the area under the graph y=x^3 using Riemann Sum. Even in high school, I have had a hard time grasping how to find the area using Riemann Sum. I do not think it was explained to me right the first time.

    Problem: find the area underneath the graph y=x^3 using eight subdivisions and rectangles underneath y.

    The range is from 0 to 1


    in my prof's example (of y=x^2) he used the following information:
    the area of each rectangle is i^2/n^3 where i=(n-1)

    so: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3


    and because: [m(m+1)]/2=1+2+3...m
    and 1^2+2^2...m^2=m(m+1)(2m+1)/6

    then: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3 = (n-1)(n)2n-1)/

    I know that the above is confusing (at least in IMO) but I really would appreciate any guidance at all. Let me know if left any information out or was not clear enough
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  2. #2
    MHF Contributor

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    Quote Originally Posted by br1tt204 View Post
    I would appreciate any help I can get with this problem! My prof usually gives us a few practice problems we should know. One of them this week was finding the area under the graph y=x^3 using Riemann Sum. Even in high school, I have had a hard time grasping how to find the area using Riemann Sum. I do not think it was explained to me right the first time.

    Problem: find the area underneath the graph y=x^3 using eight subdivisions and rectangles underneath y.

    The range is from 0 to 1
    In my prof's example (of y=x^2) he used the following information:
    the area of each rectangle is i^2/n^3 where i=(n-1)
    Okay, he is one end of the subinterval as "x" for the entire interval: if the length of the entire interval is 1, as it is in your problem, and we divide it into n subintervals, then each subinterval has length 1/n and the endpoint of the i^{th} interval is given by x= i/n. The height of the rectangle is f(x)= f(i/n)= i^2/n^2. The base of the rectangle is the length of the subinterval, 1/n, so the area is "base times height" (1/n)(i^2/n^2)= i^2/n^3.

    ("where i= (n-1)" is incorrect. i goes from 0 to n-1. n-1 is the final value of i.)

    so: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3
    and this is equal to [0^2+ 1^2+ 2^2+ ...+ (n-1)^2]/n^3


    and because: [m(m+1)]/2=1+2+3...m
    Not relevant here.

    and 1^2+2^2...m^2=m(m+1)(2m+1)/6
    This is the formula you use.

    then: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3 = (n-1)(n)2n-1)/
    [0^2+ 1^2+ 2^2+ ...+ (n-1)^2]/n^3= [(n-1)((n-1)+1)(2(n-1)+1)]/6n^3= (n-1)n(2n-1)/6n^3. you dropped the denominator.

    I know that the above is confusing (at least in IMO) but I really would appreciate any guidance at all. Let me know if left any information out or was not clear enough
    Okay, since you are dividing the interval from 0 to 1 into 8 subintervals, each subinterval has length 1/8 and the " i^{th}" interval has left endpoint i/8. At x= i/8, f(x)= x^3= (i/8)^3= i^3/8^3.

    Each rectangle has base 1/8 and height i^3/8^3, it has area (1/8)(i^3/8^3)= i^3/8^4. Add those up: i= 0 to 7. The sum can be written as
    0/8^4+ 1/8^4+ 2^3/8^4+ 3^3/8^4+ 4^3/8^4+ 5^3/8^4+ 6^3/8^4+ 7^3/8^4= [0+ 1+ 8+ 27+ 64+ 125+ 216+ 343]/4096. Now it's just arithmetic.
    Last edited by mr fantastic; November 23rd 2009 at 01:45 AM. Reason: Added some quote tags, fixed some latex tags.
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  3. #3
    Newbie br1tt204's Avatar
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    Thank you so so much! That was it!
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