Problem 88 - Thomas Calculus

Question

By using a substitution, prove that for all positive numbers \(x\) and \(y\) $$\int_{x}^{x y} \frac{1}{t} d t=\int_{1}^{y} \frac{1}{t} d t$$

Step-by-Step Solution

Answer

Using substitution \(u = \frac{t}{x}\), we transform the first integral to match the second, proving their equality.

Step 1: Understand the Problem

We need to prove the equality of two integrals: \(\int_{x}^{xy} \frac{1}{t} dt\) and \(\int_{1}^{y} \frac{1}{t} dt\), using a substitution method.

Step 2: Choose a Substitution

To simplify the integral \(\int_{x}^{xy} \frac{1}{t} dt\), we can use the substitution \(u = \frac{t}{x}\). This substitution will change the limits of integration and simplify the integrand.

Step 3: Apply Substitution

With the substitution \(u = \frac{t}{x}\), we have \(t = xu\) and the differential \(dt = x \, du\). Substituting these into the integral gives: \(\int_{x}^{xy} \frac{1}{t} dt = \int_{1}^{y} \frac{1}{xu} x du = \int_{1}^{y} \frac{1}{u} du.\)

Step 4: Simplify and Compare

Now the integral is transformed to \(\int_{1}^{y} \frac{1}{u} du\), which is exactly the same as the second integral \(\int_{1}^{y} \frac{1}{t} dt\). Hence, the two integrals are equivalent.

Key Concepts

Substitution Method

The substitution method is a clever technique often used in calculus to simplify complex integrals. In this method, we replace a complex expression with a simpler one, making it easier to integrate. This works well when you recognize a function and its derivative within the integral.

Here's how it works:

• First, we identify a substitution that can simplify the integral. In this exercise, we use the substitution \( u = \frac{t}{x} \).
• Next, express all terms in terms of the new variable \( u \). For our substitution, it means that \( t = xu \) and \( dt = x \, du \).
• Replace the original variable and limits of integration with those in terms of \( u \). This transforms the integral into a form that is often easier to evaluate.

Using substitution effectively requires practice, but it can be a powerful tool in finding integrals. It's particularly useful for turning complicated algebra into simple calculus!

Definite Integrals

Definite integrals are a fundamental concept in calculus. They allow us to calculate the total accumulation of quantities and provide a way to evaluate the area under a curve. In a definite integral, you integrate a function over a specific interval.

For example, the integral \( \int_{x}^{xy} \frac{1}{t} dt \) represents the area under the curve of \( \frac{1}{t} \) from \( x \) to \( xy \). The result of this integral gives us a specific value that corresponds to this interval.

• With definite integrals, we have limits of integration which specify the start and end points on the x-axis.
• The outcome of evaluating a definite integral provides a numerical value rather than a function, which is typical for indefinite integrals.

Using definite integrals, we can solve real-world problems involving continuous change, like calculating distance, area, and even probabilities.

Integral Equivalence

Integral equivalence refers to the idea that two different-looking integrals result in the same value. In calculus, this occurs when different approaches or substitutions lead to the same definite integral value.

Let's consider the problem again with our substitution method. We transformed the integral \( \int_{x}^{xy} \frac{1}{t} dt \) into \( \int_{1}^{y} \frac{1}{u} du \). Despite the integrals starting differently, they end up being equivalent because:

• The substitution properly adjusts the limits of integration, ensuring all parts match in value.
• The adjustment in the variable accounts for changes in scale, thus leaving the area under the curve unchanged.

This equivalence is useful, especially when checking the correctness of performed integrations or when trying to simplify the calculation of an integral!