Question

How do you find the Riemann sum of `f(x)=x^3` , -1<=x<=1, if the partition points are -1, -0.5, 0, 0.5, 1 and the sample points are -1, -0.4, 0.2, 1?

Answer 1

There are `n=4` subintervals, each of length `Delta x = 0.5`, use `sum_1^4 f(x_(i)"*") Delta x`

Explanation:

So we need:

`f(-1)*0.5 + f(-0.4)*0.5 + f(0.2)*0.5 + f(1)*0.5`

`= [f(-1) + f(-0.4) + f(0.2) + f(1)]*0.5`

`= [(-1)^3 + (-0.4)^3 + (0.2)^3 + (1)^3]*0.5`

`= [-1 - 0.064 + 0.008 + 1]*0.5`

`= (-0.56)(0.5) = -0.28`