Problem 59 - Calculus Early Transcendentals: Pearson New International Edition

Question

Decide whether the given statement is true or false. Then justify your answer. If \(\int_{a}^{b} f(x) d x>\int_{a}^{b} g(x) d x\), then $$ \int_{a}^{b}[f(x)-g(x)] d x>0 $$

Step-by-Step Solution

Answer

True. The integral of the difference is positive.

Step 1: Understanding the Premise

We are given two definite integrals: \( \int_{a}^{b} f(x) \ dx \) and \( \int_{a}^{b} g(x) \ dx \). The problem states that the integral of \( f(x) \) over \([a, b]\) is greater than that of \( g(x) \). Thus, \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \).

Step 2: Express the Difference

We need to compare the integral of the difference between the two functions. So consider the integral \( \int_{a}^{b} [f(x) - g(x)] \, dx \). This expression represents the integral of the difference between \( f(x) \) and \( g(x) \).

Step 3: Apply the Property of Integrals

Using the property of integrals \( \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \). This is a result of the linearity of integrals, which allows us to separate the integrals of the summed function.

Step 4: Determine the Result

Given that \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \), we subtract \( \int_{a}^{b} g(x) \, dx \) from \( \int_{a}^{b} f(x) \, dx \). Thus, \( \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx > 0 \).

Key Concepts

Definite Integral

The concept of a definite integral is fundamental to calculus and involves calculating the accumulation of quantities, such as areas under curves, over a specific interval \[ [a, b] \]. This notation \( \int_{a}^{b} f(x) \, dx \) signifies the definite integral of a function \( f(x) \) from \( a \) to \( b \). Essentially, it adds up infinite infinitesimal quantities to give us a finite number.
The definite integral can represent various physical and geometric ideas:

• Area: The most common magnitude is the area between the graph of the function and the x-axis, over the interval \( [a, b] \).
• Accumulation: It can also show how a quantity accumulates over time if \( f(x) \) represents a rate.
• Balance: In practical terms, it can be used to find balances if the total input and output functions are known.

In contrast to an infinite area or sum, definite integrals offer a way to sum continuously by way of limiting processes through the function's behavior over the interval.

Linearity of Integrals

The linearity of integrals is a property that significantly simplifies computation. This property essentially states that the integral of a sum of functions is equal to the sum of the integrals of each function. Mathematically, it can be presented as:

• \( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \)
• \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \)

This means it's possible to split complex integrals into simpler parts, improve efficiency in finding the integral of a difference between two functions, or scaling integrals. In our exercise, the linearity property allows for breaking down the integral \( \int_{a}^{b} [f(x) - g(x)] \, dx \) into separate parts, simplifying the comparison of \( f(x) \) and \( g(x) \) across \( [a, b] \). This property is vital for practical computations and theoretical proofs alike.

Function Comparison

In calculus, comparing definite integrals involves understanding how two functions behave over a specified range \( [a, b] \). If \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \), it indicates that the cumulative effect of \( f(x) \) over \[ [a, b] \] is greater than that of \( g(x) \).
This comparison gives insight into various situations, like:

• Evaluating the balance of forces or scenarios in physics.
• In finance, understanding which investment grows more over an interval.
• In statistics, comparing probability distributions.

The concept is used in our exercise to assert that if the integral of \( f(x) \) exceeds that of \( g(x) \), then inherently, \( \int_{a}^{b} [f(x) - g(x)] \, dx > 0 \). This result stems from function comparison principles when considered over a continuous range.

Mathematical Proofs

Mathematical proofs are like the backbone of mathematics, providing the means to verify and validate claims through logical reasoning.
In the context of this exercise, proving \( \int_{a}^{b}[f(x)-g(x)] \, dx>0 \) involves understanding:

• The premise that \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \). This is the assumption serving as the starting point.
• The application of linearity of integrals to rewrite the equation.
• Logical deduction from known properties to derive the conclusion.

Proofs often entail starting from what's given, applying known principles, and arriving at an indisputable claim. It's a methodical process that not only verifies assumptions but also builds a strong foundation for further exploration and understanding in calculus and other areas of mathematics.