Question

How do you find the Riemann sum for `f(x) = x - 2 sin 2x` on the interval [0,3] with a partitioning of n = 6 taking sample points to be the left endpoints and then the midpoints?

Answer 1

See explanation.

Explanation:

On the interval `[0,3]` with `n=6`, we get:

`Delta x = (b-a)/n = (3-0)/6 = 1/2`

The subintervals are (start at `0` and successively add `1/2`)

`[0, 1/2]` `[1/2, 1]` `[1, 3/2]` `[3/2, 2]` `[2, 5/2]` `[5/2, 3]`

Left endpoints are

`0` `" "` `1/2` `" "` `1` `" "` `3/2` `" "` `2` `" "` `5/2`

Now do the arithmetic:

`f(0)*1/2+f(1/2)*1/2+f(1)*1/2+f(3/2)*1/2+f(2)*1/2+f(5/2)*1/2`

or `[f(0)+f(1/2)+f(1)+f(3/2)+f(2)+f(5/2)]*1/2`
That is the Riemann sum for left endpoints.

Midpoints

The subintervals are (start at `0` and successively add `1/2`)

`[0, 1/2]` `[1/2, 1]` `[1, 3/2]` `[3/2, 2]` `[2, 5/2]` `[5/2, 3]`

The midpoints of the subintervals are:

`1/4` `" "` `3/4` `" "` `5/4` `" "` `7/4` `" "` `9/4` `" "` `11/4`

(Note that the differences in midpoints are also `1/2`, so after finding the first, successively add `2/4` until we have all 6 midpoints.)

So the Riemann sum for midpoints is found by doing the arithmetic:

`f(1/4)*1/2+f(3/4)*1/2+f(5/4)*1/2+f(7/4)*1/2+f(9/4)*1/2+f(11/4)*1/2`

or

`[f(1/4)+f(3/4)+f(5/4)+f(7/4)+f(9/4)+f(11/4)]*1/2`