Riemann Sum with subintervals/partition

  1. So I missed a class and am trying to figure out a question in my textbook but am completely lost. It goes a little something like this:
    Let f(x)=x3 and let P=<-2,0,1,3,4> be a partition of [-2,4].
    a) Compute Riemann Sum S(f,P*) if the points <x1*,x2*,x3*,x4*>=<-1,1,2,4> are embedded in P.

    Now I know how to calculate other Riemann Sums but I have not encountered one with a partition and subintervals yet. I tried to do the autodidactic thing and look up examples and videos but I could not find one similar to this. If I could get some help on how I approach this type of question that would be great.

    The answer is 79.

    Thanks :)
     
  2. jcsd
  3. Zondrina

    Zondrina 2,000
    Homework Helper

    You want to compute the Riemann sum. I'll give you a hand here.. Recall that [itex]Δx_i = \frac{b-a}{n}[/itex]

    What do you get when you compute this : [itex]\sum_{i=1}^{n} f(x_i)Δx_i[/itex]

    Where xi is an arbitrary point in the i'th subinterval.
     
  4. SammyS

    SammyS 9,048
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The specified partition does not have equal length sub-intervals.

    Δx1 = 2 = Δx3 .

    Δx2 = 1 = Δx4 .
     
  5. HallsofIvy

    HallsofIvy 40,946
    Staff Emeritus
    Science Advisor

    Okay, -1 lies in the interval [-2, 0], 1 lies in [0, 1] or [1, 3], take your pick, 2 lies in [1, 3] so we better pick [0, 1] for the 1, and 4 lies in [3, 4].
    The length of [-2, 0] is 0-(-2)= 2, the length of [0, 1] is 1- 0= 1, the length of [1, 3] is 3- 1= 2, and the length of [3, 4] is 4- 3= 1.

    So the Riemann sum is f(-1)(2)+ f(1)(1)+ f(2)(2)+ f(4)(1) = -13(2)+ 13(1)+ 23(2)+ 43(1)

    That's impossible- every Riemann Sum requires a partition. And, it happens, there are NO "subintervals" given unless you mean that the partition itself breaks the interval [-1, 4] into "subintervals". But, again, that is true of every Riemann Sum.

     
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