Question

Suppose that \(\mathrm{f}\) and \(\mathrm{g}\) are continuous on \([\mathrm{a}, \mathrm{b}]\), \(a \neq b\), and that \(\int_{a}^{b}(f(x)-g(x)) d x=0 .\) Show that \(f(x)=g(x)\) atleast once in \([a, b]\)

Short answer

Question: Show that if two continuous functions \(f\) and \(g\) are defined in the interval \([a, b]\) with \(a \neq b\) and satisfy the equation \(\int_{a}^{b}(f(x)-g(x)) dx = 0\), then \(f(x)=g(x)\) at least once in the interval \([a, b]\). Answer: Using the Mean Value Theorem for Integrals, we showed that there exists a point \(c \in [a, b]\) where the difference function \(h(x) = f(x) - g(x)\) is equal to zero (i.e. \(h(c) = 0\)). Therefore, at point \(c\), we have \(f(c) - g(c) = 0\), which implies that \(f(c) = g(c)\). Thus, there exists at least one point in the interval \([a, b]\) where \(f(x)=g(x)\).

Step-by-step solution

Define the difference function

Let's define a new function \(h(x)\) as the difference between the functions \(f(x)\) and \(g(x)\): $$ h(x) = f(x) - g(x) $$ This new function \(h(x)\) is continuous on \([\mathrm{a}, \mathrm{b}]\), as it is the difference of two continuous functions.

Apply Mean Value Theorem for Integrals

We are given: $$ \int_{a}^{b}(f(x)-g(x)) dx = \int_{a}^{b} h(x) dx = 0 $$ According to the Mean Value Theorem for Integrals, if \(h(x)\) is continuous on \([a, b]\), then there exists at least one point \(c\) in \((a, b)\) such that: $$ h(c)(b - a) = \int_{a}^{b} h(x) dx $$

Find the point where h(c) equals zero

Since we are given that \(\int_{a}^{b} h(x) dx = 0\), using the result from Step 2, we have: $$ h(c)(b - a) = 0 $$ Since \(a \neq b\), we know that \((b - a) \neq 0\). Therefore, the only way for the above equation to hold is if: $$ h(c) = 0 $$

Show that f(x) = g(x) at c

From Step 1, we have that \(h(x) = f(x) - g(x)\). Therefore, at point \(c\), we have: $$ h(c) = f(c) - g(c) = 0 $$ This implies that: $$ f(c) = g(c) $$ Thus, we have shown that there exists at least one point \(c \in [a, b]\) where \(f(x)=g(x)\).

Key concepts

Integral Calculus

Integral calculus is like a mathematical tool for calculating the total size or value, such as an area or a volume, by summing up tiny pieces. Imagine you're trying to figure out how much paint you need to cover a pattern on the floor. By breaking up the pattern into small, manageable sections, adding up the paint needed for each, and considering how these pieces fit together, you get the overall amount of paint you’ll need. That’s sort of what integral calculus does — it ‘adds up’ functions to find the whole under certain conditions.

In the given exercise, the integral calculus concept comes into play as we consider the function representing the difference between two continuous functions over an interval. We're summing tiny differences over that interval. The crucial thing to remember is that an integral does not just give us a static value; it provides deep insight into the nature of functions within a specified range, as in the example where the integral told us the functions must be equal at least somewhere within the interval.

Continuous Functions

Now, let's talk about continuous functions, which are smooth curves without breaks, jumps, or sharp corners. It’s like drawing a line without lifting the pencil, making them predictable and nice to work with. In mathematical terms, a function is continuous on an interval if you can trace its graph from start to end within that interval without lifting your pencil.

In our exercise, both functions f(x) and g(x) are continuous, which tells us that their difference, h(x), will also be continuous. This continuous nature of h(x) is what allows us to apply particular theorems like the Mean Value Theorem for Integrals. Without continuity, we lose this powerhouse theorem—a core principle of calculus.

Fundamental Theorem of Calculus

The fundamental theorem of calculus is like a bridge that connects two major concepts in calculus: integration and differentiation. It tells us that if we take the derivative of an integral or integrate a derivative, under the right conditions, we're basically undoing one process with the other, returning to the original function or value.

The theorem has two parts. One part allows us to evaluate integrals by identifying antiderivatives, and the other part, which directly applies to our exercise, helps us link the value of integrals to the average values of functions. Specifically, it’s the part that lets us claim there is a point c where the functions in the exercise are equal. It ties the idea of summing up tiny slices (integration) to finding the rate at which something changes at a single point (differentiation). Applied to our scenario, it gives us the ability to unveil the hidden point where f(x) and g(x) must touch, based purely on the fact that their averaged differences over an interval add up to zero.