Question

An exact integration formula Evaluate \(\int_{a}^{b} \frac{d x}{x^{2}},\) where \(0

Short answer

Based on the step-by-step solution provided, the definite integral of the function \(\int_a^b \frac{dx}{x^2}\) can be evaluated as \(\frac{1}{a} - \frac{1}{b}\) by simplifying the Riemann sum and finding the limit as n approaches infinity.

Step-by-step solution

Show the inequality \(x_{k-1}\leq\sqrt{x_{k-1}x_k}\leq x_k\)

Consider the sequence of partition points \(\left\{x_0, x_1, \ldots, x_n\right\}\) on the interval \([a, b]\). Since it is a partition, we know that \(x_{k-1} < x_k\) for \(k=1,2,\ldots,n\). Taking the square root of both \(x_{k-1}\) and \(x_k\), we obtain \(\sqrt{x_{k-1}} < \sqrt{x_k}\). Multiplying these inequalities by \(\sqrt{x_{k-1}}\) and \(\sqrt{x_k}\), respectively, results in: \(x_{k-1} \leq \sqrt{x_{k-1} x_k} \leq x_k\).

Derive a formula for \(\frac{1}{x_{k-1}} - \frac{1}{x_k}\)

We are given that \(\Delta x_k = x_k - x_{k-1}\). We want to find an expression for \(\frac{1}{x_{k-1}} - \frac{1}{x_k}\) in terms of \(x_{k-1}\), \(x_k\), and \(\Delta x_k\). To do this, we first find a common denominator for the given expression: $$\frac{1}{x_{k-1}} - \frac{1}{x_k} = \frac{x_k - x_{k-1}}{x_{k-1}x_k} = \frac{\Delta x_k}{x_{k-1}x_k}.$$

Simplify the general Riemann sum using \(x_k^* = \sqrt{x_{k-1} x_k}\)

The general Riemann sum for \(\int_{a}^{b} \frac{dx}{x^2}\) is given by $$S_n = \sum_{k=1}^{n} \frac{1}{(x_k^*)^2} \Delta x_k.$$ We are given that \(x_k^*=\sqrt{x_{k-1}x_k}\), so we substitute this into the Riemann sum to obtain $$S_n = \sum_{k=1}^{n} \frac{1}{(x_{k-1}x_k)} \Delta x_k.$$ From step 2, we know that $$\frac{\Delta x_k}{x_{k-1}x_k} = \frac{1}{x_{k-1}} - \frac{1}{x_k}.$$ Therefore, we can rewrite our sum as $$S_n = \sum_{k=1}^{n} \left(\frac{1}{x_{k-1}} - \frac{1}{x_k}\right).$$

Evaluate the definite integral \(\int_{a}^{b} \frac{dx}{x^2}\)

Now let's find the limit of our simplified sum as \(n \to \infty\): $$\int_{a}^{b} \frac{dx}{x^2} = \lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{x_{k-1}} - \frac{1}{x_k}\right).$$ We observe that most of the terms in this telescoping series cancel out, leaving only the first and the last terms: $$\int_{a}^{b} \frac{dx}{x^2} = \lim_{n\to\infty} \left(\frac{1}{a} - \frac{1}{x_1} + \frac{1}{x_1} - \frac{1}{x_{2}}+\dots+ \frac{1}{x_{n-1}}-\frac{1}{b}\right) = \frac{1}{a} - \frac{1}{b}.$$ Thus, the given definite integral evaluates to \(\int_{a}^{b} \frac{dx}{x^2} = \frac{1}{a} - \frac{1}{b}\).

Key concepts

Riemann sum

Riemann sums are a fundamental concept in integral calculus, helping us understand definite integrals. In essence, a Riemann sum approximates the area under a curve by dividing it into small rectangles. These rectangles are summed up to get a total area approximation. Mathematically, we express it as:

• The interval is divided into subintervals, each having width \( \Delta x_k \).
• For each subinterval, we take a sample point \( x_k^* \) to calculate the function value, forming the height.
• The area of each rectangle is then \( f(x_k^*) \Delta x_k \), where \( f(x) \) is the function.

The sum of all these rectangles is the Riemann sum, \( S_n = \sum_{k=1}^{n} f(x_k^*) \Delta x_k \). As the subintervals get smaller, this sum gets closer to the actual area, leading us to the definite integral.

telescoping series

A telescoping series is a unique mathematical series in which many terms cancel out when summed, leaving only a few to determine the series' value. In our example, when we simplify the Riemann sum:

• We find that the expression \( \frac{1}{x_{k-1}} - \frac{1}{x_k} \) leads to successive cancellation.
• As you add up the series, the "middle" terms cancel out, simplifying the sum.

This cancellation leads us to the ultimate evaluation of the integral, where most intermediate values disappear. For the integral \( \int_{a}^{b} \frac{dx}{x^2} \), after canceling terms, only \( \frac{1}{a} - \frac{1}{b} \) remains in the final result.

partition of interval

Partitioning an interval is crucial for calculating definite integrals using Riemann sums. Here, the interval \([a, b]\) is divided into smaller segments or subintervals:

• The points \( \{x_0, x_1, \ldots, x_n\} \) define these subintervals.
• Each subinterval \([x_{k-1}, x_k]\) has a width \( \Delta x_k = x_k - x_{k-1} \).

This division allows us to approximate the function's behavior more accurately in each small segment. The finer the partition, the more accurate the approximation of the integral becomes. This method forms the basis for understanding how integration works by summing up these small segments to approach the area under a curve.

integral calculus

Integral calculus is the branch of mathematics dealing with integrals and their properties. It focuses on finding the total size or value, such as areas under curves, volumes, and accumulated quantities. The definite integral, denoted \( \int_{a}^{b} f(x) \, dx \), is defined through limits:

• It gives the total accumulation of \( f(x) \) from \( a \) to \( b \).
• The connection between the indefinite integral and antiderivatives allows us to reverse differentiate.
• Definite integrals can also represent physical quantities like distance, area, and more.

By understanding integral calculus, we bridge the gap between the geometric representation of an area and its numerical evaluation using algebraic methods. Its concepts extend deeply into many fields, demonstrating the integral's versatility and foundational role in mathematics.