Problem 41 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int \frac{2 y d y}{y^{2}-25} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Integral Form

The given integral is \( \int \frac{2y \, dy}{y^2 - 25} \). We notice that the denominator can be factored as \((y-5)(y+5)\).

Step 2: Set up Partial Fraction Decomposition

To solve the integral using partial fractions, express \( \frac{2y}{y^2 - 25} \) as \( \frac{A}{y-5} + \frac{B}{y+5} \).

Step 3: Solve for Coefficients A and B

Set the equation \( 2y = A(y+5) + B(y-5) \). Expanding and matching coefficients gives two equations: \( 2 = A + B \) and \( 0 = 5A - 5B \). Solving these, we find \( A = 1 \) and \( B = 1 \).

Step 4: Rewrite the Integral

Substitute back into the integral: \( \int \left( \frac{1}{y-5} + \frac{1}{y+5} \right) dy \).

Step 5: Integrate Each Term Separately

Integrate \( \int \frac{1}{y-5} dy = \ln|y-5| \) and \( \int \frac{1}{y+5} dy = \ln|y+5| \).

Step 6: Combine and Simplify

The combined integral is \( \ln|y-5| + \ln|y+5| + C \), which simplifies to \( \ln| (y-5)(y+5)| + C \).

Step 7: Final Result

The evaluated integral is \( \ln|y^2 - 25| + C \).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus that involves finding the integral of a function, which can give us the area under a curve. There are various methods and strategies to approach integration problems. One such approach is "Partial Fraction Decomposition." This technique is particularly useful when dealing with rational functions.

• It involves breaking down a complex fraction into simpler fractions that are easier to integrate.
• It is crucial when the numerator's degree is less than the denominator's, allowing us to express the integral as a sum of more manageable parts.
• Factorizing the denominator first can help set up partial fractions efficiently.

In this exercise, we use partial fraction decomposition because the denominator is a quadratic polynomial, which can be factored further. Once simplified, each part is integrated separately, simplifying the process.

Logarithmic Integration

Logarithmic integration comes into play when integrals involve terms like \( \int \frac{1}{x-a} \, dx \), leading to logarithmic functions as solutions. In particular:

• When we integrate \( \frac{1}{x-a} \), the result is \( \ln|x-a| \).
• This method is vital in many calculus problems where part of the function behaves like a simple rational fraction.

In the given solution, after partial fraction decomposition, each resulting fraction was integrated individually using logarithmic integration. This approach simplifies to finding the natural logarithm of the absolute value of the linear expressions in the denominator, leading us to our result: \( \ln|y^2 - 25| + C \).

Rational Functions

Rational functions are expressions that consist of a ratio of two polynomials. They can often appear complicated, but they fall into manageable parts with the right techniques.

• If the degree of the numerator is less than the denominator, it's a proper rational function, often suitable for partial fraction decomposition.
• When integrals involve rational functions, identifying simpler components through factorization and decomposition is key.
• This allows easier integration of each piece using well-known formulas like the logarithmic integration technique discussed earlier.

In the exercise at hand, we dealt with the rational function \( \frac{2y}{y^2-25} \), where the denominator's factorization paved the way to breakdown into linear terms. This breakdown made the integral approachable and straightforward to evaluate.

Mathematics Education

Teaching mathematics, especially concepts like calculus, requires a clear and structured approach. Concepts like integration and partial fraction decomposition are critical and should be taught in a way that demystifies their complexity.

• Focusing on practical applications and step-by-step breakdowns helps students anchor abstract concepts.
• Integrating visual aids and real-life examples can further enhance understanding.
• Exercises should progress from simple to complex, allowing learners to build confidence progressively.

For students tackling this integral, it's essential to recognize the transformation from an initially complex expression to simpler integrable components. Emphasizing the rationale behind each step encourages a deeper comprehension that extends beyond procedural knowledge.