Problem 94 - 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. $$6 x=A(x+1)+B(x-2)$$

Step-by-Step Solution

Answer

A = 2, B = 4.

Step 1: Setup the Rational Function

We start with the given rational function: \[\frac{6x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}\]Simplifying the right side to a common denominator \((x+1)(x-2)\), we get:\[\frac{A(x-2) + B(x+1)}{(x+1)(x-2)}\]This implies:\[6x = A(x-2) + B(x+1)\]

Step 2: Rewrite the Equation

Expand and simplify the expression on the right-hand side:\[A(x-2) + B(x+1) = Ax - 2A + Bx + B = (A + B)x + (-2A + B)\]This means the equation becomes:\[6x = (A + B)x + (-2A + B)\]

Step 3: Create the System of Equations

We can equate the coefficients from both sides of the equation:1. The coefficient of \(x\) gives us: \[A + B = 6\]2. The constant term comparison gives: \[-2A + B = 0\]

Step 4: Solve the System of Equations

Solve the system of equations:1. From \(A + B = 6\), we get \(B = 6 - A\).2. Substitute \(B = 6 - A\) into \(-2A + B = 0\), we have: \[-2A + (6 - A) = 0\] Simplifying gives us: \[-3A + 6 = 0\]3. Solving for \(A\), we find: \[-3A = -6\] \[A = 2\]4. Substitute \(A = 2\) back into \(B = 6 - A\): \[B = 6 - 2\] \[B = 4\]

Step 5: Conclusion: Values of A and B

Thus, the values of \(A\) and \(B\) that satisfy the equation \(6x = A(x+1) + B(x-2)\) are \(A=2\) and \(B=4\).

Key Concepts

Rational Functions

Rational functions are expressions that can be written as the ratio of two polynomials. They take the general form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). Understanding these functions is essential in calculus, especially when dealing with integration. Calculus often requires simplifying complex rational functions into simpler components using a technique known as partial fraction decomposition.
Partial fraction decomposition involves breaking down a complicated rational function into a sum of simpler fractions. For instance, the function \( \frac{1}{x^2 + x - 2} \) can be decomposed as \( \frac{A}{x-1} + \frac{B}{x+2} \). By expressing the function this way, it becomes easier to integrate, as each component fraction is simpler to handle.
When we perform the decomposition, the focus is on finding the values of the constants (like \( A \) and \( B \)) that make the decomposed form equivalent to the original rational function. This often leads to setting up and solving a system of equations, which is another crucial concept in mathematics.

Systems of Equations

A system of equations is a collection of two or more equations that are solved simultaneously. In the context of partial fraction decomposition, systems of equations arise when determining the constants in the decomposed form of a rational function. Each equation in the system corresponds to a specific alignment of terms in the decomposition.
To solve these systems, we typically align matching terms on both sides of the equation. For example, with the rational function \( \frac{6x}{(x+1)(x-2)} \), after decomposition, we equate coefficients from both sides to form two equations: \( A + B = 6 \) and \(-2A + B = 0 \).
The solution of these equations involves solving them like any linear system. Start by substituting one equation into the other to isolate one variable, and then use its value to find the other variable. This method allows us to find the exact constants needed for the decomposition, making integration possible.

Calculus Integration

Calculus integration is a fundamental concept in calculus, which involves finding the antiderivative of a function. Integrating rational functions is a typical scenario where partial fraction decomposition shines. By breaking the rational function into simpler fractions, the integration process becomes a lot more manageable.
For each partial fraction, the integration is straightforward and often corresponds to basic integral types you would learn early in calculus. For example, the integral of \( \frac{A}{x-a} \) is \( A \ln|x-a| \), and knowing these basic forms is crucial to successfully apply integration to decomposed rational functions.
By integrating each partial fraction separately, you piece together the integrals to form the complete solution to the original integral problem. This technique simplifies otherwise cumbersome integral calculations into a series of standard integrals. Doing so not only aids in finding solutions but also enriches understanding of how calculus and algebra interconnect.