Integral Mean Value Theorem: Why G(x) Sign Matters

Let's dive into the Integral Mean Value Theorem and unravel why the condition that g(x) doesn't change sign on the interval $[a, b]$ is so crucial. This theorem, a cornerstone of calculus, essentially guarantees the existence of a point within an interval where the value of a function equals its average value over that interval, weighted by another function.

Understanding the Integral Mean Value Theorem

Before we get into the nitty-gritty of why g(x) must maintain its sign, let's quickly recap the theorem itself. The Integral Mean Value Theorem states that if f(x) and g(x) are continuous functions on the closed interval $[a, b]$, and g(x) does not change sign on this interval, then there exists a point c in $(a, b)$ such that:

$$\int_{a}^{b} f(x)g(x) dx = f(c) \int_{a}^{b} g(x) dx$$

In simpler terms, the integral of the product of two functions, f(x) and g(x), over the interval $[a, b]$ is equal to the value of f(x) at some point c within the interval multiplied by the integral of g(x) over the same interval. This theorem is incredibly useful in various areas of mathematics, physics, and engineering, allowing us to simplify complex integrals and approximate function values.

The Critical Role of g(x) Maintaining Its Sign

Now, let's address the heart of the matter: Why is it necessary for g(x) to maintain its sign on the interval $[a, b]$? The answer lies in how we prove the theorem. The proof typically involves constructing a new function and applying the Extreme Value Theorem or the Intermediate Value Theorem. When g(x) doesn't change sign, we can treat it as a weighting function that is either always non-negative or always non-positive.

Think of g(x) as a filter. If g(x) is always positive, it emphasizes the positive parts of f(x) in the integral. If it's always negative, it flips f(x) and emphasizes the originally negative parts. However, if g(x) changes sign, it starts canceling out the contributions from different parts of f(x) in a way that makes it impossible to guarantee the existence of that special point c.

To illustrate, imagine g(x) oscillating rapidly between positive and negative values. In this case, the integral $\int_{a}^{b} f(x)g(x) dx$ could be very small, close to zero, even if f(x) and g(x) are large in magnitude. This makes it difficult to relate the integral to a specific value of f(x) at a point c in the interval. The weighting becomes inconsistent, and the theorem breaks down.

Proof Explanation Expanded

To provide a more concrete explanation, let's outline a typical proof of the Integral Mean Value Theorem and see where the sign of g(x) becomes essential.

1. Define M and m: Let M and m be the maximum and minimum values of f(x) on the interval $[a, b]$, respectively. Since f(x) is continuous on a closed interval, the Extreme Value Theorem guarantees that these values exist.

2. **Consider the case where g(x) ≥ 0: Assume that g(x) is non-negative on $[a, b]$. Then, we have:

   $$m \int_{a}^{b} g(x) dx ≤ \int_{a}^{b} f(x)g(x) dx ≤ M \int_{a}^{b} g(x) dx$$

   This inequality holds because m ≤ f(x) ≤ M for all x in $[a, b]$, and since g(x) is non-negative, multiplying by g(x) preserves the inequality.

3. Divide by the integral of g(x): If $\int_{a}^{b} g(x) dx > 0$, we can divide through by it:

   $$m ≤ \frac{\int_{a}^{b} f(x)g(x) dx}{\int_{a}^{b} g(x) dx} ≤ M$$

   This step is valid because we are dividing by a positive number, preserving the inequality. If $\int_{a}^{b} g(x) dx = 0$, then $\int_{a}^{b} f(x)g(x) dx = 0$, and the theorem holds trivially.

4. Apply the Intermediate Value Theorem: Since f(x) is continuous and the expression $\frac{\int_{a}^{b} f(x)g(x) dx}{\int_{a}^{b} g(x) dx}$ lies between the minimum and maximum values of f(x), the Intermediate Value Theorem guarantees that there exists a point c in $(a, b)$ such that:

   $$f(c) = \frac{\int_{a}^{b} f(x)g(x) dx}{\int_{a}^{b} g(x) dx}$$

   Rearranging this equation gives us the Integral Mean Value Theorem:

   $$\int_{a}^{b} f(x)g(x) dx = f(c) \int_{a}^{b} g(x) dx$$

   If g(x) were allowed to change sign, the initial inequality m \int_{a}^{b} g(x) dx ≤ \int_{a}^{b} f(x)g(x) dx ≤ M \int_{a}^{b} g(x) dx would not necessarily hold. The positive and negative parts of g(x) could cause the integral $\int_{a}^{b} f(x)g(x) dx$ to fall outside the interval $[m \int_{a}^{b} g(x) dx, M \int_{a}^{b} g(x) dx]$, invalidating the application of the Intermediate Value Theorem.

What if we change the condition?

Now, let's consider what happens if we try to relax the condition that g(x) does not change sign. Suppose we allow g(x) to change sign on $[a, b]$. In this case, the proof outlined above breaks down, and the theorem no longer holds in general.

The fundamental issue is that the inequality $m \int_{a}^{b} g(x) dx ≤ \int_{a}^{b} f(x)g(x) dx ≤ M \int_{a}^{b} g(x) dx$ is no longer guaranteed to be true. When g(x) changes sign, the integral $\int_{a}^{b} f(x)g(x) dx$ can take on values that are outside the range of $m \int_{a}^{b} g(x) dx$ and $M \int_{a}^{b} g(x) dx$. This is because the positive and negative contributions of g(x) can interfere with the bounds established by m and M.

Counterexample

To illustrate this with a counterexample, let's consider the following functions and interval:

• f(x) = x
• g(x) = x
• $[a, b] = [-1, 1]$

Here, g(x) changes sign on the interval $[-1, 1]$. Now, let's compute the relevant integrals:

$$\int_{-1}^{1} f(x)g(x) dx = \int_{-1}^{1} x^2 dx = \frac{x^3}{3} \Big|_{-1}^{1} = \frac{2}{3}$$

$$\int_{-1}^{1} g(x) dx = \int_{-1}^{1} x dx = \frac{x^2}{2} \Big|_{-1}^{1} = 0$$

If the Integral Mean Value Theorem held, there would need to be a c in $(-1, 1)$ such that:

$$\int_{-1}^{1} f(x)g(x) dx = f(c) \int_{-1}^{1} g(x) dx$$

$$\frac{2}{3} = f(c) \cdot 0$$

$$\frac{2}{3} = 0$$

This is clearly a contradiction. Therefore, there is no such c that satisfies the Integral Mean Value Theorem when g(x) changes sign.

Alternative Conditions?

Could we replace the condition with something else? Possibly, but it would likely involve more complex restrictions. For instance, we might need to consider the number of times g(x) changes sign or impose conditions on the magnitude of g(x) when it's positive versus when it's negative. However, such conditions would likely make the theorem less practical and harder to apply.

Final Thoughts

In conclusion, the requirement that g(x) does not change sign in the Integral Mean Value Theorem is not an arbitrary one. It's a critical condition that ensures the validity of the proof and the existence of the point c. Without this condition, the theorem breaks down, as demonstrated by our counterexample. While it might be possible to formulate alternative conditions, the simplicity and elegance of the original theorem make it a cornerstone of integral calculus. So, next time you're using the Integral Mean Value Theorem, remember to check that g(x) maintains its sign – it's not just a technicality; it's the key to making the theorem work!