Question

Find the value(s) of \(c\) guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval. $$ f(x)=5-\frac{1}{x}, \quad[1,4] $$

Short answer

The value of \(c\) is the solution of the equation derived in Step 2, which is computed in Step 3. It is the only value within the interval [1,4] that makes the function \(f(x)\) equal to its average value over this interval.

Step-by-step solution

Compute the Average Value

First, calculate the average value of the function on the interval [1,4]. The average of a function over an interval \([a,b]\) is computed as \(avg = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). Thus, calculate the integral of the function \(f(x)=5-\frac{1}{x}\) over the interval [1,4] and multiply by \(\frac{1}{4-1}\) to get the average value of the function in the given interval.

Set Up the Equation Based on the Theorem

According to the Mean Value Theorem for Integrals, there exists at least one value of \(c\) in the interval [1,4], such that \(f(c)\) is equal to the average value of the function over the interval. So, the task is to solve the equation \(f(c) = avg\) for \(c\). This involves replacing \(f(c)\) with \(5-\frac{1}{c}\) and equating that to the average value computed in Step 1.

Solve for c

To solve for \(c\), isolate \(c\) on one side of the equation. First, add \(\frac{1}{c}\) to both sides to get rid of the negative sign from the right side, then multiply both sides by \(c\) to get rid of the fraction. Solve the resulting equation to find the value of \(c\).

Key concepts

Average value of a function

When we talk about the average value of a function in calculus, we're trying to find a single number that represents the entire function's behavior over a certain interval. Imagine taking every single point on the function's graph within that interval and blending them to find the middle ground. That's the average value.

To express this mathematically, we use the formula: \

\[\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]

where \(a\) and \(b\) define the interval over which we are computing the average, and \(f(x)\) is the function in question. In the example of \(f(x)=5-\frac{1}{x}\) over the interval \([1,4]\), the average value is found by integrating from 1 to 4 and then dividing by the length of the interval, which is 3 in this case. The integral is the total area under the curve of \(f(x)\), and when divided by the interval length, it yields the average height of the function, representing its average value.

Definite integrals

The concept of definite integrals is at the heart of understanding areas and accumulated quantities in calculus. When we compute a definite integral, we are finding the exact area under the curve of a function between two points, \(a\) and \(b\).

Mathematically, the notation \

\[\int_{a}^{b} f(x) \, dx\]

signifies the process of adding up infinitely small pieces of area beneath the function \(f(x)\) from point \(a\) to point \(b\). This process is known as integration. In practice, calculating a definite integral often requires using fundamental theorems of calculus and various techniques such as substitution, integration by parts, or employing known integrals of common functions. In the case of the problem provided, integral calculus allows us to find the total area under the curve of \(f(x)=5-\frac{1}{x}\) from \(x=1\) to \(x=4\), setting up the groundwork to solve for the specific value of \(c\) in the Mean Value Theorem for Integrals exercise.

Calculus Theory

The framework of calculus is built on two foundational pillars: differential and integral calculus. These two halves of calculus enable mathematicians and scientists to describe systems undergoing change and to accumulate quantities, respectively.

Within the realm of integral calculus lies the Mean Value Theorem for Integrals. This theorem serves as a bridge between algebra and calculus, indicating that for a continuous function over a closed interval, there's at least one point where the function's value is exactly equal to the function's average value over that interval.

To use the theorem effectively, one must understand the behavior of functions, how to calculate areas under curves (definite integrals), and the implications of continuity and differentiability. The exercise provided showcases an application of calculus theory: it brings together the concepts of average function value, definite integration, and the search for specific points that satisfy the conditions of the theorem. This is a prime example of how calculus theory is not merely abstract but is applied in finding concrete solutions to mathematical problems.