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Magazine Chapter 7: Problem 41
Evaluate the integrals in Exercises \(39-56\) $$ \int \frac{2 y d y}{y^{2}-25} $$
Short Answer
Expert verified
The integral evaluates to \( \ln|y^2 - 25| + C \).
Step by step solution
01
Identify the Integral Form
The given integral is \( \int \frac{2y \, dy}{y^2 - 25} \). We notice that the denominator can be factored as \((y-5)(y+5)\).
02
Set up Partial Fraction Decomposition
To solve the integral using partial fractions, express \( \frac{2y}{y^2 - 25} \) as \( \frac{A}{y-5} + \frac{B}{y+5} \).
03
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 \( B = 1 \).
04
Rewrite the Integral
Substitute back into the integral: \( \int \left( \frac{1}{y-5} + \frac{1}{y+5} \right) dy \).
05
Integrate Each Term Separately
Integrate \( \int \frac{1}{y-5} dy = \ln|y-5| \) and \( \int \frac{1}{y+5} dy = \ln|y+5| \).
06
Combine and Simplify
The combined integral is \( \ln|y-5| + \ln|y+5| + C \), which simplifies to \( \ln| (y-5)(y+5)| + C \).
07
Final Result
The evaluated integral is \( \ln|y^2 - 25| + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
