Problem 35 - Calculus Single Variable

Question

Use the information to determine \(\int_{a}^{b} f(x) d x\) and \(\int_{a}^{b} g(x) d x\). $$ \int_{a}^{b}(f(x)+2 g(x)) d x=-7, \int_{a}^{b}(g(x)-f(x)) d x=4 $$

Step-by-Step Solution

Answer

The values are \( \int_{a}^{b} f(x) \, dx = -5 \) and \( \int_{a}^{b} g(x) \, dx = -1 \).

Step 1: Understand the Given Integrals

We are given two equations involving integrals: 1. \( \int_{a}^{b}(f(x)+2g(x)) \, dx = -7 \) 2. \( \int_{a}^{b}(g(x)-f(x)) \, dx = 4 \). Our goal is to find \( \int_{a}^{b}f(x) \, dx \) and \( \int_{a}^{b}g(x) \, dx \).

Step 2: Assign Variables to Integrals

Let \( F = \int_{a}^{b}f(x) \, dx \) and \( G = \int_{a}^{b}g(x) \, dx \). This converts the integral equations into two linear equations:1. \( F + 2G = -7 \) 2. \( G - F = 4 \).

Step 3: Solve the System of Equations

To solve the system, add the two equations:1. \( F + 2G = -7 \) 2. \( G - F = 4 \)Add the above equations to eliminate \( F \):\[ (F + 2G) + (G - F) = -7 + 4 \]Simplify the equation:\[ 3G = -3 \]So, \( G = -1 \).

Step 4: Substitute to Find F

With \( G = -1 \), substitute into the first equation:\( F + 2(-1) = -7 \)\( F - 2 = -7 \)So, \( F = -5 \).

Step 5: Verify the Solution

Check these values in the second equation:\( (-1) - (-5) = 4 \) confirms correctness.Thus, the calculations verify that \( F = -5 \) and \( G = -1 \) are the solutions.

Key Concepts

Linear Equations

Linear equations are foundational tools in algebra. They express a relationship where each term is either a constant or the product of a constant and a single variable. It's like balancing a scale where the weights on both sides should be equal. In our exercise, we converted the integral expressions into the following linear equations:

• \( F + 2G = -7 \)
• \( G - F = 4 \)

These equations involve constants \(-7\) and \(4\) on one side, and expressions involving \(F\) and \(G\) on the other side. The goal is to find the values of \(F\) and \(G\) that satisfy both conditions simultaneously. Understanding linear equations involves recognizing that their graphical representation would be lines on a coordinate plane. The intersection of these lines represents the solution that satisfies both equations.

System of Equations

The term 'system of equations' refers to a set comprising two or more equations that all share the same set of variables. In this exercise, our system is comprised of two equations:

• \( F + 2G = -7 \)
• \( G - F = 4 \)

Solving a system means finding the values of the shared variables that make each equation true. This process often involves algebraic methods like substitution or elimination.
In our solution, we used the elimination method. By adding the two equations together, we cleverly eliminate \( F \), making it easier to solve for \( G \). Once \( G \) was determined, it could be substituted back into one of the original equations to solve for \( F \). This highlights the elegance and efficiency of using systems of equations in calculus problem solving, allowing us to handle multiple relationships at once.

Calculus Problem Solving

Calculus problem solving often involves integrals, which are used to find quantities like areas under curves or total accumulations. In our exercise, the task was to find definite integrals \( \int_{a}^{b} f(x) \, dx \) and \( \int_{a}^{b} g(x) \, dx \). The exercise provides complex combinations of these integrals:

• \( \int_{a}^{b} (f(x) + 2g(x)) \, dx = -7 \)
• \( \int_{a}^{b} (g(x) - f(x)) \, dx = 4 \)

The core of calculus problem solving lies in breaking down these problems into manageable parts, often using algebraic methods. By converting the integral expressions into linear equations, we apply the principles of both calculus and algebra to find the individual integrals.
Definite integrals and their properties, such as linearity, are critical in simplifying these expressions. The linearity of integrals allows us to handle each term separately within the equations, making it a flexible and powerful tool in calculus problem solving.