Question

Prove: If \(f(x)\) and \(g(x)\) are continuous for \(a \leq x \leq b\) and \(|g(x)-f(x)| \leq \epsilon\) for \(a \leq x \leq b\), then $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right| \leq \epsilon(b-a) $$ [Hint: Use (4.6).]

Short answer

Question: Prove that if two functions \(f(x)\) and \(g(x)\) are continuous on the interval \([a, b]\) and their absolute difference is at most \(\epsilon\), then the absolute difference of their integrals over this interval is at most \(\epsilon(b - a)\). Answer: We can prove this by following these steps: 1. Start with the absolute difference expression of their integrals: $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right| $$ 2. Combine the integrals to get: $$ \left|\int_{a}^{b} (g(x) - f(x)) dx\right| $$ 3. Use the given inequality \(|g(x) - f(x)| \leq \epsilon\) to obtain: $$ \left|\int_{a}^{b} (g(x) - f(x)) dx\right| \leq \int_{a}^{b} \epsilon dx $$ 4. Evaluate the integral of epsilon to get: $$ \int_{a}^{b} \epsilon dx = \epsilon(b - a) $$ 5. Conclude the proof by showing the inequality: $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right| \leq \epsilon(b - a) $$

Step-by-step solution

Start with the absolute difference expression of their integrals

We want to find an upper bound of the absolute difference of the integrals of \(f(x)\) and \(g(x)\). So, we start with the expression $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right| $$

Combine the integrals

As the integrals have the same limits of integration, we can combine them into a single integral: $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right|= \left|\int_{a}^{b} (g(x) - f(x)) dx\right| $$

Use the inequality of the difference

Now, we use the given inequality \(|g(x) - f(x)| \leq \epsilon\) in our integrals: $$ \left|\int_{a}^{b} (g(x) - f(x)) dx\right| \leq \int_{a}^{b} |g(x) - f(x)| dx \leq \int_{a}^{b} \epsilon dx $$

Evaluate the integral of epsilon

As \(\epsilon\) is a constant, we can evaluate the integral on the right hand side: $$ \int_{a}^{b} \epsilon dx = \epsilon(b - a) $$

Conclude the proof

Now, we can present the result: $$ \left|\int_{a}^{b} g(x) d x-\int_{a}^{b} f(x) d x\right| \leq \epsilon(b - a) $$ This concludes the proof.

Key concepts

Continuous Functions

Continuous functions are at the heart of calculus and play a vital role in establishing the properties of integrals. A function is said to be continuous on a closed interval \( [a, b] \) if it does not have any 'jumps' or 'breaks' within that range. This means for any value \( x \) within \( [a, b] \) the function will have a well-defined output, and small changes in \( x \) will lead to small changes in the function value \( f(x) \)..

Understanding continuity is crucial when dealing with integrals because the Fundamental Theorem of Calculus requires a function to be continuous on the interval of integration. If a function fails to be continuous, the behavior of its integral can be unpredictable or undefined, leading to more complex analysis that goes beyond standard integration techniques.

A key characteristic of continuous functions is their ability to be integrated over a closed interval, as demonstrated in the exercise we are considering. This forms the foundation of many theorems and principles in calculus.

Absolute Difference of Integrals

The absolute difference of integrals is a measure of how much two functions differ from each other when their values are accumulated over an interval. It's given by the expression:
\[\left|\int_{a}^{b} g(x) \, dx - \int_{a}^{b} f(x) \, dx\right|\].

This expression shows the absolute value of the difference between the integrals of two functions \( g(x) \) and \( f(x) \) from \( a \) to \( b \) and signifies the total 'distance' between the areas under the curves of these functions over the given interval.

In our exercise, we take advantage of the property that the integral of a non-negative continuous function represents the area under its curve. As such, if we know that the difference \( |g(x) - f(x)| \) is bounded by some value \( \epsilon \) over \( [a, b] \) then the integral of this difference, hence the absolute difference of integrals, is also bounded. This leads us to a practical application where we can estimate the maximum variation between the areas under two functions, a concept fundamental to proving our initial exercise.

Integral Evaluation

Integral evaluation involves calculating the value of an integral, which can often illuminate the relationships or differences between functions within a certain interval. In the context of the exercise provided, integral evaluation is the final step that solidifies the proof. After combining the integrals of \( g(x) \) and \( f(x) \) and using the given inequality, we find ourselves evaluating an integral with a constant integrand \( \epsilon \)..

The key to evaluating an integral of a constant over a closed interval \( [a, b] \) is to multiply the constant by the length of the interval, \( b - a \) in this case. The evaluation process confirms that the area under a constant function over an interval is simply the product of the constant and the interval's width:
\[\int_{a}^{b} \epsilon \, dx = \epsilon(b - a)\].

Through the lens of this problem, understanding how to evaluate such integrals highlights the geometric interpretation of integrals and the importance of area calculations in analysis. This step is fundamental as it reinforces the concept that the integral of a bounded function over a finite interval is also bounded—a profound concept that has far-reaching implications in all areas of calculus.