Understanding the Integral of a Derivative and the Mean Value Theorem

The question asks about the value of the definite integral of a function \(f\) over an interval \([a, b]\), where \(f\) is the derivative of some function on that interval. This scenario connects the concepts of differentiation, integration, and theorems like the Fundamental Theorem of Calculus and the Mean Value Theorem.

Let's consider the properties given:

• \(f\) is the derivative of some function on \([a, b]\). Let this function be \(F\). So, \(F'(x) = f(x)\) for all \(x \in [a, b]\).
• The function \(F\) is differentiable on \([a, b]\). A differentiable function is also continuous. Thus, \(F\) is continuous on \([a, b]\).
• Since \(f = F'\) and \(F\) is differentiable on \([a, b]\), \(f\) is defined on \([a, b]\). If \(F\) is differentiable on \([a, b]\), then \(f\) must be continuous on \([a, b]\). The Mean Value Theorem for Integrals requires the function being integrated to be continuous on the closed interval \([a, b]\).

Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (Part 2) states that if \(F\) is an antiderivative of \(f\) on \([a, b]\) (i.e., \(F'(x) = f(x)\)), then the definite integral of \(f\) from \(a\) to \(b\) is given by:

\(\int_a^b f(x) dx = F(b) - F(a)\)

This gives us the value of the integral in terms of the original function \(F\).

Applying the Mean Value Theorem for Derivatives

The Mean Value Theorem for Derivatives states that if a function \(F\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some number \(c\) in \((a, b)\) such that:

\(F'(c) = \frac{F(b) - F(a)}{b - a}\)

Since \(F\) is differentiable on \([a, b]\), it is continuous on \([a, b]\) and differentiable on \((a, b)\). Also, we know that \(F'(x) = f(x)\). So, we can rewrite the Mean Value Theorem for \(F\) as:

\(f(c) = \frac{F(b) - F(a)}{b - a}\)

for some number \(c\) in \((a, b)\).

Connecting the Theorems

We have two important relationships:

1. From the Fundamental Theorem of Calculus: \(\int_a^b f(x) dx = F(b) - F(a)\)
2. From the Mean Value Theorem for Derivatives: \(F(b) - F(a) = f(c) \cdot (b - a)\) for some \(c \in (a, b)\)

By substituting the second equation into the first, we get:

\(\int_a^b f(x) dx = f(c) \cdot (b - a)\)

This is also precisely the statement of the Mean Value Theorem for Integrals, which says if \(f\) is continuous on \([a, b]\), then there exists a number \(c\) in \((a, b)\) such that \(\int_a^b f(x) dx = f(c) \cdot (b - a)\). The condition that \(f\) is the derivative of a function on \([a, b]\) implies that \(f\) is continuous on \([a, b]\), thus satisfying the condition for the MVT for Integrals.

Evaluating the Options

Let's compare our result with the given options:

The expression \(f(c) \cdot (b - a)\) matches the result obtained from the relationship between the integral and the function's value at a specific point \(c\) as described by the Mean Value Theorem.

Revision Table: Key Calculus Theorems

Additional Information: Average Value of a Function

The expression \(\frac{1}{b-a} \int_a^b f(x) dx\) represents the average value of the function \(f\) over the interval \([a, b]\). The Mean Value Theorem for Integrals states that for a continuous function \(f\) on \([a, b]\), there exists a point \(c\) in \((a, b)\) where the function's value \(f(c)\) is equal to its average value over the interval. That is, \(f(c) = \frac{1}{b-a} \int_a^b f(x) dx\). Rearranging this gives \(\int_a^b f(x) dx = f(c) \cdot (b-a)\), which is what the question describes.

Understanding the connection between the integral (total accumulation or area) and the average value is crucial. The integral represents the "sum" of function values over the interval, and dividing by the length of the interval \((b-a)\) gives the average height or value of the function.