Problem 93 - Precalculus

Question

In calculus, when integrating rational functions, we decompose the function into partial fractions. This technique involves the solution of systems of equations. For example, suppose $$\begin{aligned} \frac{1}{x^{2}+x-2} &=\frac{1}{(x-1)(x+2)} \\ &=\frac{A}{x-1}+\frac{B}{x+2} \\ &=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)} \end{aligned}$$ and we want to find \(A\) and \(B\) such that \(1=A(x+2)+B(x-1)\) which is equivalent to \(1=(A+B) x+(2 A-B) .\) From this equation, we obtain the system of equations $$\begin{aligned} A+B &=0 \\ 2 A-B &=1 \end{aligned}$$ which solution is \(\left(\frac{1}{3},-\frac{1}{3}\right)\) Find the values of \(A\) and \(B\) that make each equation true. $$x+5=A(x+2)+B(x-4)$$

Step-by-Step Solution

Answer

The values are \(A = \frac{3}{2}\) and \(B = -\frac{1}{2}\).

Step 1: Express Given Equation

We start with the equation \(x + 5 = A(x + 2) + B(x - 4)\). Our goal is to find the constants \(A\) and \(B\) that satisfy this equality.

Step 2: Expand Right-Hand Side

Distribute \(A\) and \(B\) on the right-hand side: \[ A(x + 2) + B(x - 4) = Ax + 2A + Bx - 4B \]. Combine like terms to express it as a single equation: \(Ax + Bx + 2A - 4B\).

Step 3: Combine Like Terms

The expression becomes \((A + B)x + (2A - 4B)\). Hence, our equation is \(x + 5 = (A + B)x + (2A - 4B)\).

Step 4: Equate Coefficients

In order for the equation to hold for all \(x\), the coefficients of \(x\) and the constant terms must be equal. Thus, we form the system: \( A + B = 1 \) and \( 2A - 4B = 5 \).

Step 5: Solve the System of Equations

1. Solve the first equation for \(A\): \( A = 1 - B \).2. Substitute \(A = 1 - B\) into the second equation: \[ 2(1 - B) - 4B = 5 \] \[ 2 - 2B - 4B = 5 \] \[ 2 - 6B = 5 \]3. Rearrange to find \(B\): \[ -6B = 3 \] \[ B = -\frac{1}{2} \]4. Substitute \(B = -\frac{1}{2}\) back into \(A = 1 - B\) to find \(A\): \[ A = 1 + \frac{1}{2} = \frac{3}{2} \]

Step 6: Verify Solution

Substitute \(A = \frac{3}{2}\) and \(B = -\frac{1}{2}\) back into the equation \(x + 5 = A(x + 2) + B(x - 4)\) and simplify both sides to ensure they are equal, confirming our solution is correct.

Key Concepts

Integration of Rational Functions

Rational functions are expressions that can be written as the ratio of two polynomials. Integrating these functions involves breaking them down into simpler parts, typically using partial fraction decomposition. The general idea is that a complicated rational function can be expressed as a sum of simpler fractions that are easier to integrate.

• First, decompose the fraction – express your fraction as a sum of simpler fractions.
• Identify the constants in each partial fraction, often labeled as A, B, etc.
• Resolve the partial fractions by creating a system of equations from the original expression.

The reason for decomposing rational functions before integrating is that it simplifies the process, allowing us to integrate each smaller fraction individually. This technique is crucial in calculus for tackling complex rational expressions efficiently. Keep in mind that proper setup of your initial expression is key – factoring the denominator correctly and setting up appropriate partial fractions are the first steps towards successful integration.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system involves finding values for the unknowns that satisfy all equations simultaneously.
In our problem, we began with a rational function and decomposed it into partial fractions. To determine the values of the constants in the partial fractions, we formed a system of equations derived from equating coefficients on both sides of the equality:

• We equated the coefficients of like terms: terms involving the variable and the constant terms.
• This process generated equations like \( A + B = 1 \) and \( 2A - 4B = 5 \).
• By solving this system, we found our desired values of \( A \) and \( B \).

Dealing with systems of equations is a foundational skill in algebra and calculus. Often, solving these helps simplify and solve more complex mathematical expressions, as seen in integration.

Calculus

Calculus is the branch of mathematics that deals with continuous change, grounded in the concepts of differentiation and integration. It allows us to solve problems related to rates of change and areas under curves.
Integrating rational functions is a practical application of calculus. Here, we focus on the antiderivative, a function whose derivative returns the original function. In our context:

• Find the antiderivative of each term in the partial fraction decomposition.
• Use basic integration techniques for simpler forms like \( \frac{1}{x} \).
• Combine the results for a final integrated result.

Calculus provides the tools to address problems that involve continuous change and helps us understand the behavior of functions, including solving integrals through techniques such as partial fractions. It forms the backbone of many scientific disciplines, allowing for the analysis of dynamic systems.

Coefficients

Coefficients are numerical factors in terms of algebraic expressions. In the context of our rational function, coefficients are used to adjust and balance equations.
While dealing with partial fraction decomposition, determining these coefficients correctly is crucial. They help us ensure that:

• Each side of the equation represents the same value for any input of the variable.
• The balance of the equation, when broken into partial fractions, remains intact.
• We solve for the system of equations generated, maintaining the equality in the original problem.

Understanding how to handle coefficients is important as they link terms in algebraic expressions and are key to simplifying equations. Mastery of this concept aids in solving complex problems by breaking them down into manageable parts. Proper handling of coefficients ensures the accuracy and efficiency of solving rational integrals.