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How does one prove the midpoint rule for triple integral?

That is,

$$\sum_i \sum_j \sum_k f(\hat{x_i}, \hat{y_j}, \hat{z_k}) \Delta x_i \Delta y_j \Delta z_k$$

where $\displaystyle (\hat{x_i}, \hat{y_i}, \hat{z_i})=\bigg( \frac{x_{i-1}+x_i}{2}, \frac{y_{j-1}+y_j}{2}, \frac{z_{k-1}+z_k}{2}\bigg)$

It seems that this formula relies on that "every midpoint of every subinterval is visited".

However, I wonder if this is some kind of permutation algebra proof or whether there's some other reasoning to it?

Or does one somehow know that to get the midpoints of all of some $\mathbb{R}^n$, then midpoints of each individual $\mathbb{R}$ are composed through a product rule?

Felix Marin
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mavavilj
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  • What do you mean by prove it? – Cameron Williams Nov 11 '18 at 18:14
  • @CameronWilliams Well prove that the given formula does give the midpoint rule for the triple integral. That is, that it will approximate the integral using the midpoint rule. Surely one could otherwise believe that the given sum is just some sum, but not one that approximates a triple integral. – mavavilj Nov 11 '18 at 18:16
  • This is literally the definition of the midpoint rule. As for why it approximates the actual integral, see Darboux, I guess. – Cameron Williams Nov 11 '18 at 18:29
  • @CameronWilliams Since they're supposed to be midpoints of cubes, then how do I know that e.g. $(1/2,1/2,1)$ is such? – mavavilj Nov 11 '18 at 18:38

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