Question

True or False. Justify your answer with a proof or a counterexample. Assume all functions \(f\) and \(g\) are continuous over their domains. \(\int_{a}^{b} f(x)^{2} d x=\int_{a}^{b} f(x) d x \int_{a}^{b} f(x) d x\)

Short answer

The statement is False; the integral equality doesn't hold for all functions.

Step-by-step solution

Understand the Statement

The statement claims that the integral of the square of a function \(f(x)\) over the interval \([a, b]\) is equal to the product of the integral of \(f(x)\) over that interval, with itself. In mathematical terms, this means \(\left(\int_{a}^{b} f(x) dx \right)^2 = \int_{a}^{b} f(x)^2 dx \). We need to examine if this equality holds for all continuous functions \(f(x)\).

Consider a Simple Counterexample

Let's consider a simple function, such as \(f(x) = 1\), over the interval \([a, b]\). We will calculate both sides of the equation to see if they are equal.

Compute the Left Side for \(f(x) = 1\)

Calculate \(\left(\int_{a}^{b} 1 \cdot dx\right)^2\):\[\int_{a}^{b} 1 \cdot dx = \left[ x \right]_{a}^{b} = b-a\]Thus, the left side becomes:\((b-a)^2\).

Compute the Right Side for \(f(x) = 1\)

Calculate \(\int_{a}^{b} 1^2 \cdot dx\):\[\int_{a}^{b} 1 \cdot dx = \left[ x \right]_{a}^{b} = b-a\]Thus, the right side is simply:\(b-a\).

Compare Both Sides for \(f(x) = 1\)

For the function \(f(x) = 1\), we found that the left side \((b-a)^2\) is not equal to the right side \(b-a\), unless \(b-a = 0\). This means that the original given equation does not hold true in general as a universal statement.

Key concepts

Continuous Functions

Continuous functions form the backbone of calculus, allowing us to work with integrals smoothly. A function is considered continuous on an interval if there are no breaks, jumps, or holes in its graph within that interval. To put it simply, you should be able to draw the graph of the function over the interval without lifting your pencil. Since calculus often involves taking limits and derivatives, functions that aren't continuous can lead to complications and undefined behavior.
Why are continuous functions important? Because they guarantee that integrals behave predictably. This is crucial when working with definite integrals, which depend on the function not changing abruptly. Think of a continuous function as a reliable friend—it never surprises you with discontinuities!
In our exercise, we assume that the functions involved are continuous. This assumption is important because it allows us to apply the fundamental theorem of calculus with confidence.

Definite Integral

The definite integral is a concept that signifies the total accumulation of a quantity, determined over a specific interval \( [a, b] \). It provides a precise value representing the net area under a function's curve from \( x = a \) to \( x = b \).
The most basic idea is to consider it as a sum of infinitely small slices under the curve. Mathematicians like this because definite integrals turn the concept of an area under a curve into precise math. The integral \( \int_{a}^{b} f(x) \, dx \) gives what is called the "net area under the curve," which accounts for areas below the x-axis as negative.
For students learning integral calculus, understanding definite integrals is crucial.

• They are well-defined when the function \( f(x) \) is continuous over \( [a, b] \).
• Serve as building blocks for more advanced concepts.
• Tell you how much total "stuff" you have from \( a \) to \( b \).

This exercise involves using definite integrals to explore whether a specific equality holds.

Counterexample

A counterexample is a powerful tool in mathematics, used to demonstrate that a general statement is false. When you find a single case where the statement fails, it proves that the statement cannot be universally true across all possible scenarios. It's as if you are finding the odd one out to break an assumption.
In our exercise, to test the statement \( \left(\int_{a}^{b} f(x) \, dx \right)^2 = \int_{a}^{b} f(x)^2 \, dx \), we consider \( f(x) = 1 \). Calculating both sides with this function over the interval \( [a, b] \), we find:

• The left side becomes \( (b-a)^2 \).
• The right side simplifies to \( b-a \).

So they are only equal when \( b-a = 0 \,\text{ or } \, a = b \), which reveals that the equation doesn't hold true for all continuous functions. This simple counterexample neatly disproves the original claim.

Proof

To justify mathematical statements, we often use a proof—a logical argument that confirms the truth of the statement under certain conditions. There are many ways to prove a statement, such as direct proofs, indirect proofs, and proofs by contradiction. In the context of this exercise, using a counterexample serves as a form of proof by contradiction.
Here's why. The claim is that \( \left(\int_{a}^{b} f(x) \, dx \right)^2 = \int_{a}^{b} f(x)^2 \, dx \) for all continuous \( f \). To prove or disprove this, we decided to assume it's true and find conditions under which it doesn't hold. By showing a single function \( f(x) = 1 \) for which the statement is false, we provided concrete evidence against the claim.
So even if you were tempted to think something was true for all functions, demonstrating a single exception alters the case. In mathematics, "proof" is about showing your work and leaving no stone unturned, ensuring conclusions are drawn from solid reasoning.