Question

Show that if \(f\) and \(f^{\prime}\) are continuous for \(a \leq x \leq b\) and \(\left|f^{\prime}(x)\right| \leq K=\) const for \(a \leq x \leq b\), then for each subdivision of mesh less than \(\delta=\epsilon /[2 K(b-a)]\), each sum \(\sum f\left(x_{i}^{*}\right) \Delta_{i} x\) differs from \(\int_{a}^{b} f(x) d x\) by less than \(\epsilon\).

Short answer

Question: Prove that if \(f\) and its derivative \(f'\) are continuous on the interval \([a, b]\), and the absolute value of the derivative is bounded by a constant \(K\), then for any given \(\epsilon > 0\), there exists a partition of the interval \([a, b]\) with a mesh size smaller than \(\delta=\frac{\epsilon}{2K(b-a)}\), such that the Riemann sum of \(f\) for that partition differs from the integral of \(f\) over \([a, b]\) by at most \(\epsilon\). Answer: This statement is true because, using the Riemann-Lebesgue Lemma and considering a partition of \([a, b]\) with mesh size smaller than \(\delta=\frac{\epsilon}{2K(b-a)}\), we can show that the Riemann sum of \(f\) for that partition differs from the integral of \(f\) over \([a, b]\) by at most \(\epsilon\). This proof is achieved by applying the mean value theorem to \(f\) over the interval \([x_{i-1}, x_i]\) and deriving an inequality that demonstrates the desired difference between the Riemann sum and the integral.

Step-by-step solution

State the Riemann-Lebesgue Lemma

The Riemann-Lebesgue Lemma states that for a continuous function \(g\) on the interval \([a, b]\), given any \(\epsilon > 0\), there exists a partition \(\{x_0, x_1, ..., x_n\}\) of the interval \([a, b]\) such that the mesh size is smaller than \(\delta\) and the absolute value of the difference between the integral and the Riemann sum is less than \(\epsilon\). Mathematically, it can be written as: Given any \(\epsilon > 0\), \(\exists \delta > 0\) such that for any partition \(\{x_0, x_1, ..., x_n\}\) with mesh size smaller than \(\delta\), we have $$ \left|\sum_{i=1}^n g(x_i^*)\Delta x_i - \int_a^b g(x) dx \right| < \epsilon $$ where \(x_i^* \in [x_{i-1}, x_i]\) and \(\Delta x_i = x_i - x_{i-1}\). We will use this lemma and the given conditions on the function \(f\) and its derivative \(f'\) to derive an inequality that demonstrates the desired difference between the Riemann sum and the integral is smaller than \(\epsilon\).

Apply the mean value theorem to \(f\)

Apply the mean value theorem to the function \(f\) over the interval \([x_{i-1}, x_i]\). Since \(f\) is continuous and differentiable, there exists a number \(\xi_i\) in the interval \((x_{i-1}, x_i)\) such that $$ f'(\xi_i)=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}. $$ From the given condition, we have \(\left|f'(\xi_i)\right| \leq K\). Therefore, $$ \left|\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}\right|\leq K. $$

Derive an inequality for the difference between Riemann sum and integral

Multiply both sides of the inequality from step 2 by \(x_i - x_{i-1}\): $$ \left|f(x_i) - f(x_{i-1})\right| \leq K (x_i - x_{i-1}). \quad (1) $$ Now, consider the Riemann sum \(\sum_{i=1}^n f(x_i^*) \Delta x_i\). We have $$ \left|f(x_i^*)\Delta x_i - f(x_{i-1})\Delta x_i\right| \leq K\Delta x_i^2. \quad (2) $$ Summing the inequality (2) over all \(i\), we have $$ \left|\sum_{i=1}^n f(x_i^*)\Delta x_i - f(x_{i-1})\Delta x_i\right| \leq K\sum_{i=1}^n \Delta x_i^2. $$ Let \(S\) denote the left side of this inequality. We have $$ S \leq K\sum_{i=1}^n \Delta x_i^2. $$ Now, we bound the right side of this inequality using the given condition on the mesh size: $$ S \leq K\sum_{i=1}^n (\delta)^2 \leq K\sum_{i=1}^n \left(\frac{\epsilon}{2K(b-a)}\right)^2. $$ Finally, simplify the inequality: $$ S \leq K\sum_{i=1}^n \left(\frac{\epsilon^2}{4K^2(b-a)^2}\right)=\frac{\epsilon^2}{4(b-a)}\sum_{i=1}^n \frac{1}{(b-a)}. $$ Since \(\sum_{i=1}^n\frac{1}{(b-a)}=n\frac{1}{(b-a)}\), we have $$ S \leq \frac{\epsilon^2}{4}n. $$ Now, since \(n\leq \frac{b-a}{\delta}\), we can substitute \(\delta=\frac{\epsilon}{2K(b-a)}\) to find an upper bound on \(S\) $$ S \leq \frac{\epsilon^2}{4}\cdot\frac{2K(b-a)}{\epsilon}=\frac{\epsilon K(b-a)}{2}. $$ Then, $$ \left|\sum_{i=1}^n f(x_i^*)\Delta x_i - \int_a^b f(x) dx \right| \leq \frac{\epsilon K(b-a)}{2} < \epsilon. $$ Thus, it satisfies the statement and we have proved that each sum \(\sum f\left(x_{i}^{*}\right) \Delta_{i} x\) differs from \(\int_{a}^{b} f(x) dx\) by less than \(\epsilon\).

Key concepts

Riemann sums

In calculus, Riemann sums are a method of approximating the definite integral of a function, and they play a critical role in the definition of the integral itself. The basic idea is to divide the domain of integration, say an interval \[a, b\], into smaller subintervals, and then to sum up the areas of rectangles whose heights are determined by the function's values at specific points within these subintervals. This summation formula is typically written as \( \sum_{i=1}^n f(x_i^*)\Delta x_i \) where \(\Delta x_i = x_i - x_{i-1}\) is the width of the \(i\)-th subinterval and \(x_i^*\) is a sample point within that subinterval.

Each choice of sample points leads to a different Riemann sum, and as the width of the subintervals goes to zero, these sums can converge to the exact integral, which is symbolized as \(\int_a^b f(x) dx\). This is a direct consequence of the function being integrable, which means the function's behavior isn't too erratic over the interval of integration. By making the subintervals sufficiently small, a Riemann sum can give an approximation of the integral to any desired degree of accuracy, which sets the stage for the fundamental connection between summation of function values and integration.

Integration of continuous functions

The integration of continuous functions is a fundamental concept in analysis, precisely because continuous functions have nice properties that make them amenable to integration. A continuous function doesn't have any abrupt changes in value, which means its graph can be drawn without lifting your pen from the paper. This smoothness is what allows us to define the definite integral in terms of the area under the curve.

More formally, if \(f\) is continuous on a closed interval \[a, b\], then the function is integrable on that interval, and the integral \(\int_a^b f(x) dx\) represents the exact area under the curve of \(f\) from \(a\) to \(b\). Intuitively, integration can be thought of as an 'infinite sum' of infinitesimally small product terms \(f(x) dx\), where \(dx\) is an infinitesimally small width and \(f(x)\) is the height of an infinitesimally thin rectangle. The limit of Riemann sums, as the partition of \[a, b\] gets finer, approaches this integral. For a continuous function, this process leads to an exact value for the integral, rather than just an approximation.

Mean value theorem

The Mean Value Theorem (MVT) for integration is a cornerstone of real analysis and has important implications for understanding the behavior of functions. It states that if \(f\) is a continuous function on the closed interval \[a, b\], then there exists at least one point \(c\) within that interval such that \(f(c)\) is equal to the average value of the function on \[a, b\]. In terms of integration, this can be expressed as \(f(c) = \frac{1}{b-a} \int_a^b f(x) dx\).

The MVT essentially tells us that a continuous function on a closed interval will take on its average value at least once. This theorem is incredibly useful when we have information about a function's average rate of change over an interval but do not know its specific behavior at all points. In the context of Riemann sums, the MVT assures us that for each subinterval, we can find a representative value such that the product of this value and the width of the interval will give us a 'good enough' approximation for the area under the curve in that subsegment, which reflects the idea that as partitions get finer, the Riemann sum converges to the integral, representing the fundamental theorem of calculus linking differentiation and integration.