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Magazine Chapter 9: Problem 19
Express \(\frac{x-2}{2 x^{2}-x-3}\) in partial fractions and hence evaluate: $$ \int_{2}^{3} \frac{(x-2) \mathrm{d} x}{2 x^{2}-x-3} $$
Short Answer
Expert verified
The answer is \( -\frac{1}{2} \ln \frac{9}{7} + \ln 2 \).
Step by step solution
01
Identify the Denominator
The given fraction is \(\frac{x-2}{2x^2-x-3}\). Start by factorizing the quadratic denominator \(2x^2 - x - 3\).
02
Factorize the Quadratic Expression
Solve \(2x^2 - x - 3 = 0\). We get:\[(2x+3)(x-1) = 0\]. Therefore, \(2x^2 - x - 3\) can be written as \((2x+3)(x-1)\).
03
Set Up the Partial Fraction Decomposition
Express \(\frac{x-2}{(2x+3)(x-1)}\) as \(\frac{A}{2x+3} + \frac{B}{x-1}\).
04
Determine Values for A and B
Multiply both sides by \((2x+3)(x-1)\) to get:\[x-2 = A(x-1) + B(2x+3)\]. Expand and solve for A and B by comparing coefficients.
05
Solve the Equations for A and B
Equate coefficients of \(x\) and constant terms to get:\[A + 2B = 1\] and \[-A + 3B = -2\]. Solving these, \(A = -1\) and \(B = 1\).
06
Rewrite the Integral
Substitute \(A\) and \(B\) back into the partial fractions decomposition:\[ \int_{2}^{3} \frac{(x-2) \mathrm{d}x}{(2x+3)(x-1)} = \int_{2}^{3} \left( \frac{-1}{2x+3} + \frac{1}{x-1} \right) \mathrm{d} x \].
07
Integrate Each Term Separately
Evaluate the integral term by term:\[ \int_{2}^{3} \frac{-1}{2x+3} \mathrm{d} x + \int_{2}^{3} \frac{1}{x-1} \mathrm{d} x \].
08
Compute the Antiderivatives
The antiderivatives are:\[ -\frac{1}{2} \ln |2x+3| \] and \( \ln |x-1|\).
09
Evaluate the Definite Integral
Calculate the definite integrals:\[ \left[-\frac{1}{2} \ln |2x+3| \right]_{2}^{3} + \left[\ln |x-1| \right]_{2}^{3} \].
10
Substitute and Simplify
Substitute the limits and simplify to get:\[ -\frac{1}{2} (\ln 9 - \ln 7) + (\ln 2 - \ln 1) \]. Which simplifies to:\[ -\frac{1}{2} \ln \frac{9}{7} + \ln 2 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique where a complex rational function is broken down into simpler fractions that are easier to integrate. In this exercise, we start with \(\frac{x-2}{2x^2-x-3}\). First, we factorize the denominator into \((2x+3)(x-1)\). This allows us to express the original fraction as a sum of simpler fractions: \(\frac{A}{2x+3} + \frac{B}{x-1}\). Next, we equate coefficients to solve for constants \A\ and \B\. Simplifying these steps, we find \A = -1\ and \B = 1\. Thus, the fraction \(\frac{x-2}{(2x+3)(x-1)}\) decomposes into \(\frac{-1}{2x+3} + \frac{1}{x-1}\).
Integral Calculus
Integral calculus helps in finding the area under a curve, among other applications. To evaluate \(\frac{-1}{2x+3} + \frac{1}{x-1}\), we focus on finding the antiderivatives of each term separately. The antiderivative of \(\frac{1}{x-1}\) is \(\text{ln}|x-1|\), and for \(\frac{-1}{2x+3}\), it's \(-\frac{1}{2} \text{ln}|2x+3|\). We then apply the properties of definite integrals to solve the sum of these integrals over the interval \[2,3\].
Definite Integrals
Definite integrals evaluate the area under the curve between two specific points. After finding our antiderivatives, we apply limits 2 and 3 to each function. This results in step-by-step calculations of \[-\frac{1}{2} \text{ln}|2x+3|\] and \[\text{ln}|x-1|\] at these limits. We then substitute, simplifying to get \[-\frac{1}{2} (\text{ln}\frac{9}{7} ) + \text{ln}2\]. This final result represents the evaluated definite integral.
Quadratic Factorization
Factorizing quadratic expressions simplifies complex problems. In our case, we have \2x^2 - x - 3\. Setting it to zero, \(2x^2 - x - 3 = 0\), and solving gives us \(2x+3)(x-1)\). This factorization is crucial for partial fraction decomposition. It allows us to express our given function in simpler forms that are easier to handle in subsequent steps, especially in integration.
