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)=x^{3} and let P=<-2,0,1,3,4> be a partition of [-2,4]. a) Compute Riemann Sum S(f,P*) if the points <x_{1}*,x_{2}*,x_{3}*,x_{4}*>=<-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 :)
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 x_{i} is an arbitrary point in the i'th subinterval.
The specified partition does not have equal length sub-intervals. Δx_{1} = 2 = Δx_{3} . Δx_{2} = 1 = Δx_{4} .
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) = -1^{3}(2)+ 1^{3}(1)+ 2^{3}(2)+ 4^{3}(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.