Problem 39 - Calculus: Early Transcendental Functions

Question

Prove that if \(f\) is continuous on the interval \([a, b],\) then there exists a number \(c\) in \((a, b)\) such that \(f(c)\) equals the average value of \(f\) on the interval \([a, b]\).

Step-by-Step Solution

Answer

Following the explicit steps of integral calculus, we proved that for a function \(f\) continuous over a closed interval \([a, b]\), there always exists a number \(c\) in \((a, b)\) such that \(f(c)\) equals the average value of \(f\) on this interval \([a, b]\).

Step 1: Define the average value

First, the average value of a function \(f(x)\) over the interval \([a, b]\) is given by the formula \(A = \frac{1}{b-a} \int_a^b{f(x)dx}\)

Step 2: Define the integral function

Next, define the integral function \(F(x) = \int_a^x{f(t)dt}\) over \(x\) ranges in \([a, b]\)

Step 3: Use the integral function to prove the existence of \(c\)

By the Fundamental Theorem of Calculus, \(F(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), which means \(F'(x) = f(x)\) for all \(x\) in \((a, b)\). The function \(F(x)\) will start at \(F(a) = 0\) and end at \(F(b) = (b-a)A\). Now we create a new function \(G(x) = (b-a)Ax - f(x)\), which is also continuous on \([a, b]\) and differentiable on \((a, b)\). We can see that \(G(a) = 0\) and \(G(b) = (b-a)A - (b-a)A = 0\). Because \(G(a) = G(b)\), by the Mean Value Theorem (the version for derivatives), there must exist a point \(c\) in \((a, b)\) that \(G'(c) = 0\)

Step 4: Prove \(f(c)\) equals the average value of \(f\)

According to the mean value theorem, \(G'(c) = 0\) implies \(G'(c) = 0 = (b-a)A - f(c)\). Therefore, we found that \(f(c) = A\), which means at point \(c\), \(f(c)\) equals the average value of \(f\) on \([a, b]\).

Key Concepts

Continuous Functions

A function is called continuous if, graphically speaking, it can be drawn without lifting your pen from the paper. Mathematically, a function \( f(x) \) is continuous on an interval \( [a, b] \) if it is continuous at every point in that interval.
The continuity of a function implies several important properties:

• No sudden jumps or gaps on the interval.
• The function values must be predicted using limits.

In the context of the Mean Value Theorem for Integrals, continuity guarantees that the function behaves nicely over the entire interval, allowing us to make conclusions about its average value.

Average Value of a Function

The average value of a function over a specific interval provides us with a single number, representing the "typical" behavior of the function on that interval. For a continuous function \( f(x) \) over an interval \( [a, b] \), the average value is computed using the formula:\[A = \frac{1}{b-a} \int_a^b{f(x) \, dx}\]This formula essentially sums up all the values of the function within the interval and divides by the length of the interval \((b-a)\), giving a mean value.
It forms a foundational aspect in understanding how a function distributes its values across an interval, particularly when proving related theorems like the one in the exercise.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, showing that they are essentially inverse processes. It is divided into two parts:

• The first part states that if \( F(x) \) is an antiderivative of \( f(x) \) on \( [a, b] \), then the integral of \( f \,\text{from} \ a \ ext{to} \ b \) is \( F(b) - F(a) \).
• The second part states that if \( f \) is continuous on \( [a, b] \), and \( F \) is defined by \( F(x) = \int_a^x f(t) \, dt \), then \( F \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \), and \( F'(x) = f(x) \).

In the given solution, this theorem justifies the step where \( F'(x) = f(x) \), ensuring that both the integration and differentiation processes hold for \( f(x) \), and it aids in finding the average value.

Integral Calculus

Integral calculus is a major branch of mathematics concerned with accumulation and total change.It provides tools for calculating areas under curves, among many other uses.
The integral of a function \( f(x) \) over an interval \[ a, b \] is expressed as \( \int_a^b f(x) \, dx \). This process helps us find the total accumulation of quantities, such as areas, volumes, displacement, etc.
In this problem, integral calculus is employed to determine the average value of the function \( f(x) \) across the interval. It lets us compute \( \int_a^b f(x) \, dx \), which is crucial for understanding how the entire function behaves across the interval. By dividing this accumulated value by the interval's length, we derive the average value. This technique nicely unifies our understanding of total accumulation with average rate of change.