Question

How do you evaluate the Riemann sum for 0 ≤ x ≤ 2, with n = 4?

Answer 1

Here is the general procedure.

Explanation:

For a given `f`, with `0 ≤ x ≤ 2`, and `n = 4`

We cut the interval `[0,2]` into 4 equal subintervals (assuming a uniform partition), each of length `Delta x = (b-a)/n = (2-0)/4 = 1/2`

The intervals are:

`[0,1/2]`, `[1/2,1]`. `[1, 3/2]`, `[3/2,2]`

Now we need `x_1"*"` in `[0,1/2]`

and `x_2"*"` in `[1/2,1]`

`x_3"*"` in `[1, 3/2]`

`x_4"*"` in `[3/2,2]`

The Riemann sum will be:

`f(x_1"*") 1/2 + f(x_2"*") 1/2 +f(x_3"*") 1/2 +f(x_4"*") 1/2`

or, if you prefer:

`[f(x_1"*") + f(x_2"*") +f(x_3"*") +f(x_4"*")] 1/2`

Until some function and some convention for selecting `x_i"*"` is specified, that is the best we can do.