Problem 19 - Calculus for Biology and Medicine

Question

Let \(f(x)=2 x, 0 \leq x \leq 2 .\) Use a geometric argument to find the average value of \(f\) over the interval \([0,2]\), and find \(x\) such that \(f(x)\) is equal to this average value.

Step-by-Step Solution

Answer

The average value of \(f(x)\) over \([0, 2]\) is 2, and \(f(1) = 2\).

Step 1: Understanding the Function

We are given the function \(f(x) = 2x\) defined over the interval \([0, 2]\). This function is a linear function with a slope of 2. On the interval, the function goes from \(f(0) = 0\) to \(f(2) = 4\).

Step 2: Calculate the Average Value

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by \( \frac{1}{b-a} \int_{a}^{b} f(x)\, dx\). For our function, this becomes \( \frac{1}{2-0} \int_{0}^{2} 2x\, dx\).

Step 3: Set Up the Integral

Calculate the integral \(\int_{0}^{2} 2x\, dx\). This requires finding the antiderivative of \(2x\), which is \(x^2\). Evaluate this from 0 to 2.

Step 4: Evaluate the Integral

Using the Fundamental Theorem of Calculus, evaluate \(x^2\) from 0 to 2: \([x^2]_0^2 = (2^2) - (0^2) = 4 - 0 = 4\).

Step 5: Compute the Average Value

Substitute the result of the integral into the average value formula: \( \frac{1}{2} \times 4 = 2\). So, the average value of \(f(x)\) on \([0, 2]\) is 2.

Step 6: Find x Such That f(x) Equals Average

We need to find \(x\) such that \(f(x) = 2\). Set \(2x = 2\) and solve for \(x\).

Step 7: Solve the Equation

Solving \(2x = 2\) gives \(x = 1\).

Key Concepts

Integral Calculus

Integral calculus is a powerful mathematical tool used to find areas under curves, among other things. It involves calculating the integral of a function over a specified interval. In this exercise, we are dealing with the function \(f(x) = 2x\) over the interval \([0, 2]\). To find the average value of the function over this interval, we utilize the formula for the average value of a function, which is \( \frac{1}{b-a} \int_{a}^{b} f(x)\, dx\).

This formula simplifies to \( \frac{1}{2} \int_{0}^{2} 2x\, dx\) for our particular case. Understanding the concept of integration is crucial, as it helps us to accumulate the changes represented by the function over a continuous interval.

• The integral \( \int_{0}^{2} 2x\, dx\) is solved by finding the antiderivative. For \(2x\), the antiderivative is \(x^2\).
• By applying the limits of integration, we find that \([x^2]_0^2 = 4\).

The final step connects the concept of integration to the computation of an average value, providing a comprehensive understanding of how areas under linear functions can be calculated mechanically and conceptually.

Linear Functions

Linear functions, like our function \(f(x) = 2x\), are among the simplest and most fundamental types of functions in mathematics. They create straight lines when plotted on a graph and are characterized by constant rates of change or slopes.

In our scenario, the linear function \(f(x) = 2x\) has a slope of 2, indicating that for every unit increase in \(x\), \(f(x)\) increases by 2. This type of function is described by the general form \(y = mx + b\), where \(m\) represents the slope and \(b\) the y-intercept.

• In our function, \(m = 2\) and \(b = 0\), which means the line passes through the origin (0,0) and slopes upwards twice as fast as \(x\) increases.
• Linear functions are essential in calculus because they facilitate understanding more complex, non-linear functions through linear approximation and other techniques.

Grasping the behavior of these functions helps students not only solve equations involving them but also interpret and predict real-world phenomena where change at a constant rate is involved.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the worlds of differential and integral calculus, two pillars of this fascinating mathematical field. It has two main parts: the first part relates the process of differentiation and integration, while the second part provides a way to compute exact values of definite integrals.

In our exercise, we use the second part of the theorem to evaluate the integral \( \int_{0}^{2} 2x\, dx\). Here's how it helps:

• This theorem tells us that if we can find an antiderivative, \(F(x)\), of a function \(f(x)\), the integral of \(f(x)\) over \([a, b]\) is simply \(F(b) - F(a)\).
• In our example, \(F(x) = x^2\) is the antiderivative of \(2x\).

This means we can directly calculate the integral by evaluating \(x^2\) at the boundaries of the interval, providing us with \((2^2) - (0^2) = 4\). This theorem is a fundamental tool because it simplifies the daunting task of calculating areas, making problems manageable and solvable.