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In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$ \int_{4}^{8} \frac{y d y}{y^{2}-2 y-3} $$

Short Answer

Expert verified
The evaluated integral is \( \ln(\sqrt{15}) \).

Step by step solution

01

Factor the Denominator

The given integral is \( \int_{4}^{8} \frac{y \, dy}{y^{2}-2y-3} \). Start by factoring the quadratic in the denominator, \( y^2 - 2y - 3 \). This can be factored as \( (y - 3)(y + 1) \).
02

Set Up Partial Fractions

Express \( \frac{y}{(y-3)(y+1)} \) as a sum of partial fractions. We assume \( \frac{y}{(y-3)(y+1)} = \frac{A}{y-3} + \frac{B}{y+1} \). Multiply both sides by the denominator \((y-3)(y+1)\) to clear the fractions: \( y = A(y+1) + B(y-3) \).
03

Solve for Constants

Expand and equate coefficients: \( y = Ay + A + By - 3B \). This gives \( y = (A+B)y + (A-3B) \). Comparing coefficients, we get the system: \( A+B = 1 \) and \( A-3B = 0 \). Solve these equations: \( A = \frac{3}{4} \) and \( B = \frac{1}{4} \).
04

Substitute Back into Integral

Substitute back into the integral: \( \int_{4}^{8} \left( \frac{3/4}{y-3} + \frac{1/4}{y+1} \right) dy \). Split into two integrals: \( \frac{3}{4} \int_{4}^{8} \frac{1}{y-3} dy + \frac{1}{4} \int_{4}^{8} \frac{1}{y+1} dy \).
05

Evaluate Each Integral

Each integral is of the form \( \int \frac{1}{x} dx = \ln |x| + C \). So, evaluate: \[ \frac{3}{4} \left[ \ln |y-3| \right]_{4}^{8} + \frac{1}{4} \left[ \ln |y+1| \right]_{4}^{8} \].
06

Calculate the Definite Integrals

Calculate each part: For the first integral: \( \frac{3}{4} \left( \ln|8-3| - \ln|4-3| \right) = \frac{3}{4} (\ln 5 - \ln 1) = \frac{3}{4} \ln 5 \). For the second integral: \( \frac{1}{4} \left( \ln|8+1| - \ln|4+1| \right) = \frac{1}{4} (\ln 9 - \ln 5) = \frac{1}{4} \ln \frac{9}{5} \).
07

Combine Results

Combine the results: \( \frac{3}{4} \ln 5 + \frac{1}{4} \ln \frac{9}{5} = \frac{3}{4} \ln 5 + \frac{1}{4} (\ln 9 - \ln 5) \). Simplify this: \( \frac{3}{4} \ln 5 + \frac{1}{4} \ln 9 - \frac{1}{4} \ln 5 = \frac{1}{2} \ln 5 + \frac{1}{4} \ln 9 \). Factor further to combine logs: \( \ln 5^{1/2} + \ln 9^{1/4} \). Use properties of logarithms to combine: \( \ln \left( 5^{1/2} \times 9^{1/4} \right) = \ln \left( \sqrt{5} \times \sqrt[4]{9} \right) \).
08

Final Answer

The expression simplifies to \( \ln \left( 5^{1/2} \times 3^{1/2} \right) = \ln \left( \sqrt{15} \right) \). This is the final evaluated integral.

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Key Concepts

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

Definite Integral
In calculus, a definite integral is an essential concept used to find the accumulated quantity like area under a curve within a specific interval. The integral of a function between two points, say from 4 to 8 as in the given problem, gives the definite integral. This process involves evaluating an antiderivative at the upper limit and subtracting the value of the antiderivative at the lower limit.
To solve an integral problem like this one, certainly breaking down complex expressions using techniques such as partial fraction decomposition can simplify the process. When faced with the definite integral \( \int_{a}^{b} f(x) \, dx \), it's important to account for the exact limits \(a\) and \(b\) in the calculations.
Factorization
Factorization is the process of breaking down a complex expression into simpler components, which can be multiplied to give the original expression. This process significantly simplifies integrals involving rational functions, as seen in this exercise.
To accomplish factorization, look for patterns or use formulas that help split polynomials into factors. In this exercise, the expression \( y^2 - 2y - 3 \) is factored into two parts: \( (y - 3)(y + 1) \).
This step transforms a complex fraction into manageable parts. By separating them, we can later employ partial fraction decomposition to make integration practical and straightforward.
Integral Evaluation
The integral evaluation involves solving the problem by finding the antiderivative of the function and using the definite integral limits. Once a rational function is broken down into partial fractions, each separate fraction can be integrated independently.
For the integral \( \int \frac{1}{x} \, dx \) encountered here as part of the decomposition, the antiderivative is \( \ln |x| + C \).
Therefore, given two separate integrals after partial fraction decomposition like \( \int_{4}^{8} \left( \frac{3/4}{y-3} + \frac{1/4}{y+1} \right) \, dy \), solve each piece by substituting the limits of the integral and subtracting the results as outlined in the solution.
Logarithm Properties
Logarithms play a crucial role in solving integrals that involve expressions like \( \int \frac{1}{x} \, dx \). One key property is \( \ln(ab) = \ln(a) + \ln(b) \), which helps combine or separate log terms effectively.
This exercise demonstrates the use of such properties, with logarithmic identities aiding in simplifying the expression further by factoring constants out of the logarithms. For instance, applying \( \frac{1}{4} \ln 9 - \frac{1}{4} \ln 5 \) can be rewritten by these properties into a single logarithm expression: \( \ln \left( \frac{9}{5} \right) \).
Understanding these logarithm properties ensures you can transform your final answer into a neater, condensed form as seen in the solution's conclusion.

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Most popular questions from this chapter

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