Question

Use the definition of an integral as the limit of a sum and the fact that absolute value of a sum is less than or equal to the sum of absolute values to prove the Darboux inequality.

Short answer

The Darboux inequality \(m \cdot (b-a) \leq \int_a^b f(x) dx \leq M \cdot (b-a)\) is proven using the assumption that \(m \leq f(x) \leq M\) for all \(x\) in the [a,b] interval, the definition of an integral as a limit of a sum, and the fact that the absolute value of a sum is less than or equal to the sum of absolute values.

Step-by-step solution

Understanding the Problem

The Darboux inequality is an integral inequality that states that for a function \(f:\[a,b\] \to R\) which is Riemann integrable, and \(m\), \(M\) stands for the infimum and supremum of \(f\) on \[a,b\] respectively, then the following inequality holds: \(m(b-a) \leq \int_a^b f(x) dx \leq M(b-a)\).\nTo show this, we will rely on two key principles: the definition of an integral as a limit of a sum, and the property of absolute values that says the absolute value of a sum is less than or equal to the sum of absolute values.

Working with the Definition of an Integral

By the definition of integrability, for a function \(f(x)\) we have: \(m \leq f(x) \leq M\), where \(m\) and \(M\) are the infimum and supremum of \(f\) over the interval [a, b]. We will now work on the integral part. Recall that the integral of \(f\) from a to b is defined as \( \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x\) where \(x_i\) is a point in the i-th subinterval and \(\Delta x = \frac {b- a}{n}\).

Using the Darboux Inequality

Because \(m \leq f(x_i) \leq M\) for each \(i\), we know that \(m \cdot \Delta x \leq f(x_i) \cdot \Delta x \leq M \cdot \Delta x\). Summing up these inequalities from \(i = 1\) to \(n\) , we get \(m \cdot (b-a) \leq \sum_{i=1}^{n} f(x_i) \cdot \Delta x \leq M \cdot (b-a)\). Taking limit as \(n \to \infty\), we obtain \(m \cdot (b-a) \leq \int_a^b f(x) dx \leq M \cdot (b-a)\), proving the Darboux inequality.

Key concepts

Integral as a Limit of a Sum

Understanding the concept of the integral as a limit of a sum is fundamental in calculus. The idea is to approximate the area under a curve by dividing it into small sections, which are easier to calculate. Imagine the interval \([a, b]\) on the x-axis being divided into 'n' smaller intervals of equal width, denoted by \(\Delta x\), where \(\Delta x = \frac{b-a}{n}\). For each subinterval, we take a sample point \(x_i\) and evaluate the function \(f(x_i)\) at each point.

This gives us a sum of terms \(f(x_i) \cdot \Delta x\), which approximates the integral of \(f\) from \(a\) to \(b\). As \(n\) approaches infinity, the width of these subintervals becomes infinitesimally small, making the sum converge to the exact value of the integral:\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x \]Understanding this limit process is crucial for grasping the concept of the Riemann integral, which is foundational for proving inequalities and understanding properties of functions over intervals.

Riemann Integrable Function

A Riemann integrable function is one that can be approximated by simple geometric shapes, like rectangles, to determine the area under its curve over a certain interval. Not all functions are Riemann integrable.

For a function \(f\) to be considered Riemann integrable on the interval \([a, b]\), it must meet a specific criterion: for every possible partition of the interval into subintervals, both the lower sum \(L(f, P)\) and the upper sum \(U(f, P)\) converge to the same value as the maximum width of the subintervals goes to zero. In simpler terms, the infimum and supremum of the function over every subinterval should allow the area approximation via sums to closely approach each other.

This means:

• The function must be bounded over the interval.
• It can't have too many points of discontinuity.

Functions that are continuous over a closed interval are always Riemann integrable. This property is vital when considering the applications of calculus, especially in defining and computing integrals in more complex settings.

Infimum and Supremum

The concepts of infimum and supremum are central to understanding bounds of functions, particularly in the context of the Darboux inequality. They provide a way to describe the smallest and largest bounds that a function can achieve over a specific interval.

The **infimum** of a function over an interval \([a, b]\), denoted by \(m\), is the greatest lower bound. It's the highest value that is less than or equal to every value of the function in the interval. In simpler settings, it can be thought of as the minimum value, but it doesn't necessarily have to be achieved by the function within the interval.

On the other side, the **supremum**, denoted by \(M\), is the least upper bound. It's the smallest value that is greater than or equal to every value of the function over the interval—again, not necessarily achieved by the function itself.

In applying these concepts to Darboux's inequality, we use them to constrain the function \(f\) such that:

• The area under \(f\) from \(a\) to \(b\) reflects these bounds.
• The integral of \(f\) is squeezed between the infimum and supremum times the length of the interval \(b-a\).

This insight is crucial for proving that the approximation of the integral is always flanked by these bounds, laying the foundation for various integral inequalities used in calculus.