Extremum of a function of several variables

1. The volume under the surface is given by the double integral,

We split up the integration region into a 2x3 grid of rectangles whose upper right corners are determined by the right endpoints of the partition along either axis. That is, we split up the interval [0, 1] into 2 subintervals,

[0, 1/2], [1/2, 1]

with right endpoints given by the arithmetic sequence,

for , and the interval [0, 1] into 3 subintervals,

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

with right endpoints

for .

Then the upper right corners of the 6 rectangles are the points

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

generated by the sequence .

The integral is thus approximated by the sum

or approximately 2.4334. (Compare to the actual value of the integral, which is close to 2.952.)

For the midpoint rule estimate, we replace the sampling points with , i.e. the midpoints of each subinterval, so the set of sampling points is

(1/4, 1/6), (3/4, 1/6), (1/4, 1/2), (3/4, 1/2), (1/4, 5/6), (3/4, 5/6)

and the integral is approximately

or about 2.908.

2. We approach the second integral the same way. Split up the interval into 8 subintervals with left and right endpoints given respectively by

for , and the interval into 2 subintervals with

for .

The upper left corners of the rectangles in this grid are given by the sequence . So the integral is approximately

(Compare to the actual value, 32.)

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