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Show that \(\int_{a}^{b} \frac{d t}{t}=\int_{1 / b}^{1 / a} \frac{d t}{t}\) when \(0

Short Answer

Expert verified
The integrals are equal due to substitution and reversing the integration limits after substitution gives the same result.

Step by step solution

01

Substitute in the integral

To show that two integrals are equal, we will first consider the transformation property of integrals in relation to substitution. Let's make a substitution to transform the limits of the integral on the left-hand side to match values on the right. Set \(u = \frac{1}{t}\).
02

Determine the differential substitution

Calculate the differential \(d u\) with respect to \(t\):\[d u = -\frac{1}{t^2} d t\] Then, \(d t = -t^2 d u = -\frac{1}{u^2} d u\) because \(t = \frac{1}{u}\).
03

Change limits of integration

Using the substitution \(u = \frac{1}{t}\), adjust the limits of integration. When \(t = a\), \(u = \frac{1}{a}\), and when \(t = b\), \(u = \frac{1}{b}\). So the new integral becomes:\[\int_{1/b}^{1/a} \frac{-1}{u} d u\].
04

Simplify the integral

The integral with negative sign becomes:\[\int_{1/b}^{1/a} \frac{-1}{u} d u = - \int_{1/b}^{1/a} \frac{1}{u} d u\].Reversing the limits of integration gives:\[- \int_{1/b}^{1/a} \frac{1}{u} d u = \int_{1/a}^{1/b} \frac{1}{u} d u\].
05

Compare both integrals

Now, compare the transformed integral to the right-hand side of the original equation:\[\int_{1/b}^{1/a} \frac{d t}{t}\] We have shown that the substitution yields the same integral values, confirming the equality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
Substitution in integrals is quite similar to the chain rule in differentiation. It's about manipulating the variable in the integral to simplify the overall expression. Let's break this down.
When we want to show the equality of two integrals, we often use the substitution method. Imagine replacing one variable with another to change the form of the integral. In our exercise, by setting a new variable, say, letting \( u = \frac{1}{t} \), we make several changes:
  • The integral bounds or limits change as well. The limits transform according to the substitution.
  • The differential \( dt \) changes. We express it in terms of \( du \), by differentiating our substitution equation \( t = \frac{1}{u} \).
Through substitution, complex forms become easier to handle and integrate. It's a strategic way to solve integrals with challenging limits or integrands.
Limits of Integration
The limits of integration are important in integral calculus. They set the start and finish points for evaluating the integral. When using the substitution method, these limits change along with the variable change.
In the given problem, the limits for the left integral are from \( a \) to \( b \). After substituting \( t = \frac{1}{u} \), the new limits become from \( \frac{1}{b} \) to \( \frac{1}{a} \). This is achieved by substituting the boundary values into the substitution equation.
  • When \( t = a \), substituting gives \( u = \frac{1}{a} \).
  • When \( t = b \), substituting gives \( u = \frac{1}{b} \).
Adjusting integration limits ensures the new variable maintains correct bounds, representing the same interval as the original function. This precise handling of limits maintains the equivalence between the transformed integral and the original.
Integral Equality
Integral equality can occur through correct transformations and substitutions. It confirms different integral expressions yield the same result. Achieving this involves transforming one side to match the other using substitution.
Initially, we had two integrals which looked different. By transforming using substitutions, and adjusting integration limits, we brought them to the same form. The sequence is like settling a puzzle - slowly fitting all pieces correctly until both sides match perfectly.
This requires maintaining the correct form of the expression and limits, ensuring transformation steps are meticulously managed through mathematical rules. Confirming equality demonstrates the power of integrals in connecting different mathematical expressions seamlessly.

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