Problem 39 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{2 y d y}{y^{2}-25}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |y^2 - 25| + C \).

Step 1: Rewrite the Expression

First, recognize that the integral is in the form of a rational function.\( \int \frac{2y \, dy}{y^2 - 25} \) can be rewritten by noticing that \( y^2 - 25 \) is a difference of squares, which factors as \( (y - 5)(y + 5) \). Thus, we can express the integral as \( \int \frac{2y}{(y - 5)(y + 5)} \, dy \).

Step 2: Apply Partial Fraction Decomposition

Perform partial fraction decomposition to express \( \frac{2y}{(y - 5)(y + 5)} \) as a sum of simpler fractions: \( \frac{A}{y - 5} + \frac{B}{y + 5} \). Multiply through by the denominator \( (y - 5)(y + 5) \) to get \( 2y = A(y + 5) + B(y - 5) \).

Step 3: Solve for A and B

By expanding and solving the equation : \( 2y = Ay + 5A + By - 5B \), group and compare coefficients: \( A + B = 2 \) and \( 5A - 5B = 0 \). Solving this system of equations gives \( A = 1 \) and \( B = 1 \).

Step 4: Rewrite the Integral Using Partial Fractions

Substitute \( A \) and \( B \) back into the partial fraction: \( \frac{1}{y - 5} + \frac{1}{y + 5} \). Rewrite the integral as: \( \int \left( \frac{1}{y - 5} + \frac{1}{y + 5} \right) \, dy \).

Step 5: Integrate

Now integrate each term separately: \( \int \frac{1}{y - 5} \, dy = \ln |y - 5| + C_1 \) and \( \int \frac{1}{y + 5} \, dy = \ln |y + 5| + C_2 \). Combine these to get \( \ln |y - 5| + \ln |y + 5| + C \).

Step 6: Simplify the Expression

Use the logarithmic identity \( \ln a + \ln b = \ln(ab) \): \( \ln |y - 5| + \ln |y + 5| = \ln |(y - 5)(y + 5)| \). The final solution is \( \ln |y^2 - 25| + C \) where C is the constant of integration.

Key Concepts

Rational Functions

Rational functions are a key concept in calculus, often expressed as the ratio of two polynomials. In our case, the rational function to be integrated is \( \frac{2y}{y^2 - 25} \). Understanding these functions involves recognizing characteristics, such as factoring polynomials to reveal a simpler structure. Here, the polynomial in the denominator, \( y^2 - 25 \), can be factored as a difference of squares: \((y - 5)(y + 5)\). This kind of factorization is crucial as it sets the stage for decomposition into partial fractions.To break this down further:

• The **numerator** is the polynomial \( 2y \).
• The **denominator** is the product \((y - 5)(y + 5)\), derived from factoring \( y^2 - 25 \).

This decomposition simplifies the process of integration by allowing us to express it as a sum of simpler, easily integrable fractions.

Calculus Problem-Solving

Solving problems in calculus often involves several clearly defined steps to arrive at a solution. Each step should be approached methodically to demystify the process. In tackling rational functions, particularly those involving partial fractions, this methodical approach becomes essential. Here's an overview of the typical steps:

• **Identify the form.** Recognize the type of rational function and note any factors or simplifications like differences of squares.
• **Apply partial fraction decomposition.** Break the function into simpler fractions whose denominators match factored components.
• **Solve for coefficients.** Determine the constants that allow expressing the rational function as a sum of simpler fractions.
• **Integrate each component.** Once the function is decomposed, integrate each term separately using standard antiderivative rules.

This strategic problem-solving approach ensures that complex integrals are broken down into manageable parts, facilitating an easier path to the solution.

Indefinite Integrals

Indefinite integrals are a fundamental aspect of calculus, representing the process of antiderivation. They involve finding a function whose derivative is the original function being integrated.When dealing with the integral \( \int \frac{2y}{y^2 - 25} \, dy \), once decomposed into its partial fractions, the goal is to find the antiderivatives of each resulting term. The integration of each term follows a standard pattern, often involving logarithmic functions:

• For \( \int \frac{1}{y - 5} \, dy \), the result is \( \ln |y - 5| + C_1 \).
• Similarly, \( \int \frac{1}{y + 5} \, dy \) yields \( \ln |y + 5| + C_2 \).

The final integration step involves combining these results. Using logarithmic properties, these individual antiderivatives simplify to \( \ln |y^2 - 25| + C \). Here, \( C \) represents the constant of integration, highlighting the indefinite nature of the integral. This means the integral represents a family of functions, each differing by a constant.