Question

Suppose that \(f\) is continuously differentiable on a rectangle \(R .\) Show that there is a constant \(M\) such that $$ \left|\sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\right| \leq M\|P\| $$ if \(\sigma\) is any Riemann sum of \(f\) over a partition \(P\) of \(R .\)

Short answer

Question: Prove the existence of a constant M such that the absolute difference between the Riemann sum of a continuously differentiable function f over a rectangle R and the integral of f over R is less than or equal to M times the norm of the partition P. Answer: Given that the function f is continuously differentiable, we can establish the inequality, $$ \left|\sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\right| \leq (M-m)\|P\|, $$ where \(M\) and \(m\) represent the supremum and infimum of the function on the rectangle R, respectively, and \|P\| denotes the partition norm. The constant \(M\) in this case is given by the difference between the supremum and infimum of the function.

Step-by-step solution

Define supremum and infimum of the function

Let \(m\) and \(M\) be the infimum and the supremum of \(f(\mathbf{X})\) over the rectangle \(R\), respectively.

Define Riemann sum and partition norm

Let \(\sigma = \sum_{i=1}^n f(\mathbf{X}_i^*) \Delta A_i\) be a Riemann sum of \(f\) using a partition \(P=\{A_1,A_2,\dots,A_n\}\). Here, \(\mathbf{X}_i^* \in A_i\) is a point at which \(f\) is evaluated. The partition norm \(\|P\|\) is defined as the maximum diameter of the subrectangles, i.e., \(\|P\| = \max\limits_{1\leq i \leq n} \operatorname{diam}(A_i)\).

Refine the partition

Let us refine the partition such that its norm is less than \(\delta\), where \(\delta>0\). Then, the partition \(P\) meets the condition \(m\Delta A_i \leq f(\mathbf{X}_i^*) \Delta A_i \leq M\Delta A_i\) for each subrectangle \(A_i\), because \(m \leq f(\mathbf{X}_i^*) \leq M\) for all \(\mathbf{X}_i^*\in A_i\).

Use integration properties

From the previous step, we can infer the following inequality: $$ \sum_{i=1}^n m\Delta A_i \leq \sigma \leq \sum_{i=1}^n M\Delta A_i. $$ Now, we can integrate with respect to the area to get: $$ m\int_R d\mathbf{X} \leq \sigma \leq M\int_R d\mathbf{X}. $$ Subtracting the integral \(\int_R f(\mathbf{X}) d\mathbf{X}\) from both sides, we obtain: $$ m\int_R d\mathbf{X} - \int_R f(\mathbf{X}) d\mathbf{X} \leq \sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\leq M\int_R d\mathbf{X} - \int_R f(\mathbf{X}) d\mathbf{X}. $$

Apply the mean value theorem

Since f is continuously differentiable on R, we can apply the mean value theorem for integrals, which states that if a function is continuous on a closed interval, then there exists a number c in the interval such that the integral of the function over the interval is equal to the function evaluated at c times the length of the interval. In our case, we have: $$ \int_R f(\mathbf{X}) d\mathbf{X} = f(\mathbf{c})\int_R d\mathbf{X}, $$ where \(\mathbf{c} \in R\). Thus, we can rewrite the inequality obtained in step 4 as: $$ \left|m-f(\mathbf{c})\right|\int_R d\mathbf{X} \leq \left|\sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\right|\leq \left|M-f(\mathbf{c})\right|\int_R d\mathbf{X}. $$ Divide both sides by \(\int_R d\mathbf{X}\), then we have: $$ \left|m-f(\mathbf{c})\right| \leq \left|\sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\right| \leq \left|M-f(\mathbf{c})\right|. $$ Finally, since \(\left|m-f(\mathbf{c})\right|\leq M-m\) and \(\left|M-f(\mathbf{c})\right|\leq M-m\), we can conclude that: $$ \left|\sigma-\int_{R} f(\mathbf{X}) d \mathbf{X}\right| \leq (M-m)\|P\|. $$ In this case, the constant \(M\) is given by the difference between the supremum and infimum of the function.

Key concepts

Continuously Differentiable Functions

Understanding continuously differentiable functions is fundamental when we're talking about the precision of Riemann sums in approximating integrals. A function is said to be continuously differentiable over a region, like the rectangle R mentioned in our exercise, if it has a derivative at every point inside that region, and the derivative itself is a continuous function.

Why is this important? When a function is continuously differentiable, it behaves nicely — it doesn't have any sharp corners or jumps, which means its rate of change is well-behaved and predictable. This smoothness assures us that when we approximate an integral with a Riemann sum, we won't have any surprising spikes in error due to the function jerking wildly.

Integration Properties

In the realm of calculus, integration properties are the rules that govern how integrals operate. They are crucial when trying to understand the relationship between a function and its Riemann sum approximation.

For example, one of the properties states that if a function is integrated over a larger interval, the result will be greater or equal to the integral over any subinterval. This is used in the step-by-step solution to ensure that the Riemann sum remains bounded within the supremum and infimum of f times the area of the rectangle. Understanding these properties allows us to make accurate statements about the integral, even if we're working with approximations.

Mean Value Theorem for Integrals

The mean value theorem for integrals is a guarantee that a continuous function has an average value on a closed interval. To be precise, it states that there exists at least one point c in the interval over which the function is being integrated such that the integral of the function over the interval is exactly equal to the function's value at c multiplied by the length of the interval.

This theorem is a cornerstone when proving the boundedness of the Riemann sum error because it bridges the gap between the discrete nature of Riemann sums and the smooth, continuous essence of integrals. By applying this theorem, we're essentially finding the constant M in the original exercise that bounds the difference between the Riemann sum and the true value of the integral.

Partition Norm

The partition norm, or simply the norm of a partition, is a measure of how fine or coarse a partition is. In the Riemann sum context, it refers to the length of the longest subinterval when the interval is divided into portions.

The smaller the partition norm, the finer the partition, which in turn means a more accurate approximation of the integral: the Riemann sum 'fits' the function more closely. Conversely, a larger partition norm hints at a poorer fit and potentially a larger error. By controlling the partition norm, mathematicians can control the approximation error, which is pivotal in the solution to the exercise. It ensures that as the partition gets more refined — meaning the norm gets smaller — the sum comes closer to the true value of the integral.