Problem 36 - Calculus Single Variable

Question

Use the information to determine \(\int_{a}^{b} f(x) d x\) and \(\int_{a}^{b} g(x) d x\). $$ \int_{a}^{b}(f(x)-7 g(x)) d x=-9, \int_{a}^{b}(g(x)+f(x)) d x=0 $$

Step-by-Step Solution

Answer

\(\int_{a}^{b} f(x) \, dx = -\frac{9}{8} \) and \(\int_{a}^{b} g(x) \, dx = \frac{9}{8} \).

Step 1: Express Given Integrals

We are provided with two integral equations: \( \int_{a}^{b}(f(x)-7 g(x)) \, dx=-9 \) and \( \int_{a}^{b}(g(x)+f(x)) \, dx=0 \). We need to solve these equations simultaneously to find \( \int_{a}^{b} f(x) \, dx \) and \( \int_{a}^{b} g(x) \, dx \).

Step 2: Define Variables for Integrals

Define \( I_f = \int_{a}^{b} f(x) \, dx \) and \( I_g = \int_{a}^{b} g(x) \, dx \). The integral equations can now be rewritten as:1. \( I_f - 7I_g = -9 \)2. \( I_f + I_g = 0 \)

Step 3: Solve the System of Equations

Substitute the second equation into the first. From \( I_f + I_g = 0 \), we find \( I_f = -I_g \). Substitute this into \( I_f - 7I_g = -9 \): \( -I_g - 7I_g = -9 \).This simplifies to \( -8I_g = -9 \).

Step 4: Solve for \( I_g \)

Solve for \( I_g \): \[ -8I_g = -9 \]Divide both sides by -8: \[ I_g = \frac{9}{8} \].

Step 5: Solve for \( I_f \)

Use the result \( I_g = \frac{9}{8} \) to find \( I_f \). From \( I_f = -I_g \), we have: \[ I_f = -\frac{9}{8} \].

Key Concepts

Definite Integrals

Definite integrals play a crucial role in calculus, allowing us to calculate the net area under a curve within a specific interval. In our context, we dealt with \(\int_{a}^{b} f(x) \, dx\) and \(\int_{a}^{b} g(x) \, dx\), which represent the areas under the curves of functions f(x) and g(x) from points \( a \) to \( b \).

• The integral symbol \( \int \) is followed by the function we want to integrate.
• The letters \( a \) and \( b \) are called limits of integration, indicating the interval along the x-axis we are interested in.
• If the result of a definite integral is positive, it suggests the function lies generally above the x-axis over the interval, and if negative, below the x-axis.

Understanding definite integrals is vital as it helps us to evaluate the total accumulation of quantities, not just geometric areas but also things like distances, probabilities, and other physical quantities.

In our problem, solving for these integrals revealed how multiple functions can be analyzed together over the same interval, emphasizing the power of calculus in handling complex, interconnected systems.

System of Equations

A system of equations consists of two or more equations working together, aiming to find solutions for multiple unknown variables simultaneously. This concept is central in the exercise where we are given the equations \( I_f - 7I_g = -9 \) and \( I_f + I_g = 0 \).

• Each equation in a system holds some information about the variables involved.
• By solving such a system, we aim to find values for all the variables that satisfy each equation simultaneously.

In practice, solving a system often involves techniques such as substitution or elimination. For our integral problem, the substitution method was employed: one equation expressed \( I_f \) in terms of \( I_g \), allowing substitution into the other to solve the system.

These systems underpin not just mathematics, but applications in real-world physics, engineering, economics, and countless fields where variables interact in complex ways.

Simultaneous Equations

A special case of systems of equations, simultaneous equations are solved together to find common solutions, applicable to various aspects of mathematics including our exercise on integrals. The beauty of simultaneous equations lies in their ability to handle multiple unknowns efficiently, as seen in our equations \( I_f + I_g = 0 \) and \( I_f - 7I_g = -9 \).

• Start by simplifying one of the equations, making one variable the subject.
• Substitute this into the other equation to reduce the number of variables.
• This often transforms a problem into simpler, solvable equations.

In this context, we discovered that \( I_f = -I_g \) from the simpler equation \( I_f + I_g = 0 \). This strategic move was pivotal in further cracking open the solution to the other equation.

Simultaneous equations not only serve mathematics but are cornerstones in technical fields like algorithm design, chemistry for reaction balances, and in financial modeling, proving their versatility and invaluable flexibility.