Riemann Sum with subintervals/partition

1. Jan 17, 2013

MelissaJL

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. Jan 17, 2013

Zondrina

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

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

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

3. Jan 17, 2013

SammyS

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

Δx1 = 2 = Δx3 .

Δx2 = 1 = Δx4 .

4. Jan 18, 2013

HallsofIvy

Staff Emeritus
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.