Problem 33 - Calculus Single Variable

Question

In Exercises \(33-36,\) use the information to determine \(\int_{a}^{b} f(x) d x\) and \(\int_{a}^{b} g(x) d x\). $$ \int_{a}^{b}(f(x)-3 g(x)) d x=3, \int_{a}^{b}(-6 g(x)+9 f(x)) d x=6 $$

Step-by-Step Solution

Answer

\(\int_{a}^{b} f(x) \, dx = 0\) and \(\int_{a}^{b} g(x) \, dx = -1\).

Step 1: Understand the Given Equations

We have two integrals given: \(\int_{a}^{b}(f(x)-3 g(x)) \, dx=3\) and \(\int_{a}^{b}(-6 g(x)+9 f(x)) \, dx = 6\). These are linear combinations of the integrals of \(f(x)\) and \(g(x)\). Our goal is to find the individual integrals \(\int_{a}^{b} f(x) \, dx\) and \(\int_{a}^{b} g(x) \, dx\).

Step 2: Set Variables for the Integrals

Let's define \(I_1 = \int_{a}^{b} f(x) \, dx\) and \(I_2 = \int_{a}^{b} g(x) \, dx\) such that we can express the given information as:1. \(I_1 - 3I_2 = 3\)2. \(9I_1 - 6I_2 = 6\).

Step 3: Solve the System of Equations

To isolate \(I_1\) and \(I_2\), first use the equations:1. \(I_1 - 3I_2 = 3\)2. \(9I_1 - 6I_2 = 6\).First, multiply the first equation by 9 to align the \(I_1\) coefficients:\(9I_1 - 27I_2 = 27\).Now subtract it from the second equation:\[(9I_1 - 6I_2) - (9I_1 - 27I_2) = 6 - 27\], which results in: \(21I_2 = -21\). Thus, \(I_2 = -1\).

Step 4: Substitute to Find the Other Integral

Use \(I_2 = -1\) in the first equation \(I_1 - 3I_2 = 3\):\[I_1 - 3(-1) = 3\]\[I_1 + 3 = 3\]\[I_1 = 0\]. Thus, \(\int_{a}^{b} f(x) \, dx = 0\).

Key Concepts

Systems of Equations

Systems of equations are mathematical tools that allow us to find unknown values based on given relationships. They consist of two or more equations that share the same set of variables. When solving systems of equations, our goal is to find the exact values of the unknown variables by making the given equations true simultaneously.
To solve a system of equations, there are several methods at our disposal. Here are a few common ones:

• Substitution: Solve one equation for one variable, and substitute this expression into the other equation. This simplifies the system to a single equation.
• Elimination: Manipulate the equations by adding or subtracting them to eliminate one of the variables, reducing the system to a single equation.
• Graphing: Plot both equations on a graph. The solution is the intersection point of the graphs, representing the values that satisfy both equations.

In the exercise provided, the substitution method was utilized after manipulating the coefficients to align the variables and solve them systematically. This process showcased the importance of systems of equations in simplifying complex scenarios involving multiple conditions.

Integral Calculus

Integral calculus is a branch of mathematics concerned with areas, accumulations, and the anti-derivative of functions. It involves the computation of integrals, which can be understood as the reverse process of differentiation.
Definite integrals, like the ones in the exercise, accumulate values from a starting point \( a \) to an endpoint \( b \). This is denoted by \( \int_{a}^{b} f(x) \, dx \). Integrals can describe quantities like areas under curves, total accumulated quantities, and more.
There are two main types of integrals:

• Definite Integrals: Provide a number that represents the net area under a curve between two bounds, \( a \) and \( b \). It's the integral with limits and includes both the bounds and the function \( f(x) \).
• Indefinite Integrals: Provide the general form of the antiderivative of a function, denoted without bounds.

In the given situation, we worked with definite integrals to express relationships as linear combinations, simplifying the problem to a system of equations. Understanding these integral types is crucial for solving real-world calculus problems.

Linear Algebra

Linear algebra revolves around the study of vectors, vector spaces, and linear mappings. It offers powerful techniques for solving systems of linear equations, like the ones in the exercise.
In the realm of calculus, linear algebra helps in:

• Representing systems: Any linear system can be expressed in the form of matrices, facilitating easier manipulation and solving.
• Vector spaces: Solutions to linear equations can be examined through vector spaces, understanding how different solutions relate.
• Transformations: Applying linear transformations can simplify systems, allowing for straightforward problem-solving through various matrix operations.

By using linear algebra, we can systematically address complex problems by breaking them down into more manageable parts. This was evident in the exercise as we defined the integrals as variables \( I_1 \) and \( I_2 \), culminating in a straightforward application of elimination to find solutions.