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)

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

can i see a photo of the question so i can you?