In the midpoint rule for triple integrals we use a triple riemann sum to approximate a triple integral over a box b, where f(x, y, z) is evaluated at the center (xi, yj, zk) of the box bijk. use the midpoint rule to estimate the value of the integral. divide b into eight sub-boxes of equal size. (round your answer to three decimal places.)triple integrals of cos(xyz) dv, where b = {(x, y, z) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2}

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling at the 8 points,

(1/2, 1/2, 1/2)

(1/2, 1/2, 3/2)

(1/2, 3/2, 1/2)

(3/2, 1/2, 1/2)

(1/2, 3/2, 3/2)

(3/2, 1/2, 3/2)

(3/2, 3/2, 1/2)

(3/2, 3/2, 3/2)

which are captured by the sequence

with each of being either 1 or 2.

Then the integral of over is approximated by the Riemann sum,

(compare to the actual value of about 4.159)

2 is the final ! : )