Problem 201 - Calculus Volume 1

Question

Explain why, if \(f\) is continuous over \([a, b]\) and is not equal to a constant, there is at least one point \(M \in[a, b]\) such that \(f(M)=\frac{1}{b-a} \int_{a}^{b} f(t) d t\) and at least one point \(m \in[a, b]\) such that \(f(m) < \frac{1}{b-a} \int_{a}^{b} f(t) d t\)

Step-by-Step Solution

Answer

A continuous and non-constant function attains its average value and also takes values below it.

Step 1: Define the Average Value of the Function

The average value of a continuous function \( f \) over the interval \([a, b]\) is given by the formula:\[ \frac{1}{b-a} \int_{a}^{b} f(t) \, dt. \]This value represents the mean value of the function across the interval \([a, b]\).

Step 2: Use the Mean Value Theorem for Integrals

By the Mean Value Theorem for Integrals, if \( f \) is continuous over \([a, b]\), there exists at least one point \( M \in [a, b] \) such that:\[ f(M) = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt. \]This guarantees that the function \( f \) takes on its average value at least once in the interval.

Step 3: Demonstrate the Existence of a Point Smaller Than the Average

Since \( f \) is not a constant function, it must vary over \([a, b]\). This means \( f \) will take values both below and above its average value:- There exists a point \( m \in [a, b] \) such that \( f(m) < \frac{1}{b-a} \int_{a}^{b} f(t) \, dt. \)This variation in function values is due to the continuity and non-constancy of \( f \).

Key Concepts

Continuous Function

A continuous function is one that has no breaks, jumps, or holes in its graph. You can draw it without lifting your pencil from the paper. This smooth nature of continuous functions allows us to make some important conclusions, like using the Mean Value Theorem for Integrals.

For a function to be continuous on an interval \( [a, b] \), it should not have any undefined points within that interval. This concept guarantees that every value between \( f(a) \) and \( f(b) \) is covered by the function. This quality is especially useful in calculus as it enables us to apply theorems that depend on this property.

• Continuous functions behave predictably and smoothly over an interval.
• They are essential for applying the Mean Value Theorem for Integrals.
• They ensure a function’s average value is attainable within its domain.

Understanding the continuity of a function helps in analyzing its behavior and using calculus to find specific values related to the function's mean or average.

Average Value of a Function

The average value of a function over an interval gives insight into what a 'typical' output of the function might be across that stretch. It's like averaging your grades to find out how you've been doing in class overall.

For a continuous function \( f \) over \( [a, b] \), the average value is calculated using the integral:
\[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt. \] This formula combines the integrals with division by the interval length to find a central value for the function over \( [a, b] \). The outcome is a single, representative value of the function over that interval.

• The formula involves integration and the concept of division by the interval length.
• It helps us find a representative value of \( f(t) \) over \( [a, b] \).
• This value is crucial for applying the Mean Value Theorem for Integrals.

Having a solid grasp of the average value concept can help answer questions about the function's behavior and its relationship to the integral calculus.

Integral Calculus

Integral calculus is a fundamental part of calculus focused on the concept of integration. This process accumulates quantities over an interval, often giving the total or the 'sum' of the function's area under its graph.

The definite integral \( \int_{a}^{b} f(t) \, dt \) is a vital tool in integral calculus. It calculates the net area between the function \( f(t) \) and the x-axis from \( t = a \) to \( t = b \). When divided by \( b-a \), it helps calculate the average value of the function, as previously mentioned.

• Integral calculus focuses on the summation of areas to find totals or averages.
• It uses definite integrals to analyze and compute these values over specified intervals.
• Integration thus forms the backbone of understanding cumulative behavior in continuous functions.

Knowledge of integral calculus enables a deeper understanding of how functions behave over an interval and supports applying important theorems, such as the Mean Value Theorem for Integrals.