91Ó°ÊÓ

U-Substitution is a powerful technique used to simplify the process of integration. It works by transforming a complicated integral into a simpler one by substituting part of the integral with a new variable. This can make evaluating the integral more straightforward. In the given exercise, we use the substitution \( u = \sqrt{x} \). By choosing this substitution, the expression \( \sqrt{x} \) is replaced with \( u \), which directly simplifies parts of the integral.
This change in variable makes differentiating and substituting easier, transforming the original bounds of integration. When \( x = 1 \), \( u \) becomes 1, and when \( x = 3 \), \( u \) changes to \( \sqrt{3} \).
The role of U-Substitution is to not only simplify the algebra within the integral, but also to facilitate evaluating definite integrals by adapting the bounds accordingly. This kind of detailed transformation allows students to see how algebraic manipulation can make calculations more manageable.

An indefinite integral, unlike its definite counterpart, represents a family of functions and does not calculate an area but rather an antiderivative. In the context of the exercise, we first transformed our problem using U-Substitution. The integral \( \int \frac{2}{u^2 + 1} \, du \), once evaluated, resulted in the antiderivative \( 2 \cdot \tan^{-1}(u) + C \). The \( C \) here is called the constant of integration, representing the notion that there are infinitely many antiderivatives.
The indefinite integral allows us to reformulate the derived antiderivative back into terms of the original variable by reversing the substitution. For instance, substituting back \( u = \sqrt{x} \) gave us \( 2 \cdot \tan^{-1}(\sqrt{x}) + C \).
This expression can then be used to evaluate definite integrals, as shown in the solution steps. Mastery of indefinite integrals is crucial for solving more complex calculus problems, as it provides a tool for simplifying expressions and back-substituting to solve definite integrals, connecting the concept back to the original function.

Integrating trigonometric functions often involves using known identities or transformations that simplify the integration process. In this exercise, after the substitution and simplification, the integral \( \int \frac{2}{u^2 + 1} \, du \) takes a form that can be associated with the derivative of the arctan function.
This step leverages the integral identity:

• \( \int \frac{1}{1 + u^2} \, du = \tan^{-1}(u) + C \)

As a result, integrating \( \frac{2}{u^2 + 1} \) gives \( 2 \cdot \tan^{-1}(u) + C \). Understanding the relationship between trigonometric functions and their integrals is key for many calculus problems. Knowing that \( \tan^{-1}(u) \) corresponds to our function is part of integrating trigonometric functions,
Reflecting on the function's derivatives can often lead students to the correct integral form, allowing for the exact calculation needed in evaluating indefinite or definite integrals.