Question

Show that \(\int_{a}^{b} x d x=\frac{1}{2}\left(b^{2}-a^{2}\right)\) by completing the fol- lowing argument. For the partition \(a=x_{0}

Short answer

The integral \(\int_{a}^{b} x \, dx\) equals \(\frac{1}{2}(b^2 - a^2)\).

Step-by-step solution

Understand the Partition and Midpoint

Given a partition defined by \(a = x_0 < x_1 < \cdots < x_n = b\), we select the midpoint \(\bar{x}_i = \frac{1}{2}(x_{i-1} + x_i)\). Our task is to find and simplify the Riemann sum \(R_P = \sum_{i=1}^{n} \bar{x}_i \Delta x_i\) for \(f(x)=x\).

Write Riemann Sum Terms

The Riemann sum is given by \(R_P = \frac{1}{2} \sum_{i=1}^{n} (x_i + x_{i-1})(x_i - x_{i-1})\). Notice that \(\Delta x_i = x_i - x_{i-1}\), hence \(R_P = \frac{1}{2} \sum_{i=1}^{n} (x_i^2 - x_{i-1}^2)\).

Simplify the Riemann Sum

Recognize the telescoping nature of the sum. Expanding \(\sum_{i=1}^{n} (x_i^2 - x_{i-1}^2)\) results in a sequence where successive negative and positive terms cancel each other out, leaving \(x_n^2 - x_0^2 = b^2 - a^2\).

Apply the Limit to the Riemann Sum

As the mesh of the partition approaches zero, the Riemann sum converges to the exact value of the definite integral. Thus, \(\int_{a}^{b} x \, dx = \lim_{n \to \infty} R_P = \frac{1}{2} (b^2 - a^2)\).

Conclude the Argument

The integral \(\int_{a}^{b} x \, dx\) is therefore equal to \(\frac{1}{2} (b^2 - a^2)\), as demonstrated through the simplification and limit of the Riemann sum.

Key concepts

Riemann Sum

In calculus, a Riemann Sum is a method for approximating the total value of an integral, which can be thought of as the area under a curve. To understand this, we start by dividing the interval \([a, b]\) into smaller partitions and computing the value of our function at specific points within these partitions. These specific points can be chosen in various ways, such as using left endpoints, right endpoints, or midpoints (the approach used in this exercise).
Given the partition \(a = x_0 < x_1 < \cdots < x_n = b\), the Riemann sum is given by
\[R_P = \sum_{i=1}^{n} \bar{x}_i \Delta x_i\]
where \(\bar{x}_i\) is typically the midpoint of each sub-interval and \(\Delta x_i = x_i - x_{i-1}\). The simplicity of Riemann sums makes them a foundational tool for understanding the process of integration, especially when further simplified using telescoping series.

Telescoping Series

A telescoping series is one where consecutive terms cancel each other out in a manner similar to a telescope folding together. In the Riemann sum, telescoping occurs through the expression
\[\sum_{i=1}^{n}(x_i^2 - x_{i-1}^2)\]
When expanded, the negative term from each pair cancels the positive term of the next. This results in a drastically simplified expression. In our case, it leaves us with just the initial term \(b^2\) minus the final term \(a^2\).
This property not only simplifies the computation but also demonstrates the power of recognizing patterns within mathematical sums. By converting the sum into such a form, we effectively reduced complexity, which is particularly beneficial when working within integral boundaries. This highlights the mathematical elegance and cleverness involved in dealing with infinite series and sums.

Limit of a Partition

The concept of limiting a partition is crucial to transitioning from Riemann sums to definite integrals. After simplifying the Riemann sum through telescoping, we take the limit as the mesh size approaches zero. The mesh size is defined by the largest \(\Delta x_i\) in the partition. Here, we consider an infinite number of partitions as fine as possible.
As \(n \rightarrow \infty\), the partitions become infinitesimally small, meaning the sum approaches the exact area under the curve. We mathematically express this concept as:
\[\lim_{n \to \infty} R_P = \frac{1}{2} (b^2 - a^2)\]
This limit process reflects the fundamental nature of integral calculus, which sums infinitely small products of height and width, ultimately leading to precise calculations of areas, volumes, and other quantities.

Integral Calculus

Integral calculus is the mathematical study of accumulation, by which we calculate quantities such as area, volume, and other concepts that necessitate summing infinitesimal elements. The force of integral calculus lies in its ability to provide exact solutions, known as definite integrals.
Definite integrals, such as \(\int_{a}^{b} x \, dx\), evaluate the net area between a function \(f(x)\) and the x-axis, over a specified interval \[a, b\]. This particular integral is efficiently calculated using our previous results from the Riemann sum's limit:
\[\int_{a}^{b} x \, dx = \frac{1}{2} (b^2 - a^2)\]
Integral calculus takes the foundational step of summation seen in Riemann sums to the next level, which not only approximates but precisely calculates these values. In learning integral calculus, one develops a versatile tool for diverse applications, ranging from physics to economics, yielding insights across a variety of disciplines.