Problem 35 - Calculus and its applications

Question

Find the average function value over the given interval. $$ y=4-x^{2} ; \quad[-2,2] $$

Step-by-Step Solution

Answer

The average function value over the interval \([-2, 2]\) is \( \frac{8}{3} \).

Step 1: Understand the Problem

We are asked to find the average value of the function \( y = 4 - x^2 \) over the interval \([-2, 2]\). The average value of a continuous function \( f(x) \) over an interval \([a,b]\) is calculated using the formula \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \).

Step 2: Set up the Integral

Identify \( f(x) = 4 - x^2 \) and the interval \([-2, 2]\). The length of the interval is \( b - a = 2 - (-2) = 4 \). We now set up the definite integral for the given function: \( \int_{-2}^{2} (4 - x^{2}) \, dx \).

Step 3: Calculate the Definite Integral

Evaluate \( \int_{-2}^{2} (4 - x^2) \, dx \). First find the antiderivative: \( \int (4 - x^2) \, dx = 4x - \frac{x^3}{3} + C \). Now, compute this from \(-2\) to \(2\):\[ \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \].

Step 4: Evaluate the Expression

Calculate each part:- For \( x=2 \): \( 4(2) - \frac{2^3}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \).- For \( x=-2 \): \( 4(-2) - \frac{(-2)^3}{3} = -8 + \frac{8}{3} = -\frac{24}{3} + \frac{8}{3} = -\frac{16}{3} \).Now substitute these back into the expression: \( \frac{16}{3} - (-\frac{16}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \).

Step 5: Find the Average Value

Using the formula \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \), substitute the results from the previous steps: \( \frac{1}{4} \times \frac{32}{3} = \frac{32}{12} = \frac{8}{3} \). Therefore, the average value of the function \( y = 4 - x^2 \) over the interval \([-2, 2]\) is \( \frac{8}{3} \).

Key Concepts

Definite Integral

The concept of a definite integral is crucial in calculus. It is used to compute the accumulation of quantities. For example, areas under curves or displacement over time. In the context of finding an average value of a function, the definite integral helps in determining how much "total" quantity accumulated over an interval.
In our original problem, the function is given as \( f(x) = 4 - x^2 \), and the interval of integration is from \( -2 \) to \( 2 \). The definite integral \( \int_{-2}^{2} (4 - x^2) \, dx \) signifies the net area between the curve \( y = 4 - x^2 \) and the x-axis from \( -2 \) to \( 2 \).
Here's a simplified breakdown on how to calculate such an integral:

• Identify the function you are integrating across a specific interval, in this case, \( 4 - x^2 \).
• Set up the integral: \( \int_{-2}^{2} (4 - x^2) \, dx \).
• Evaluate this integral to find the total accumulation over the interval.

The evaluated integral gives us \( \frac{32}{3} \), signifying this total value from \(-2\) to \(2\) before averaging it over the 4-unit interval.

Antiderivative

Understanding antiderivatives is foundational in calculus, especially for solving integrals. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). Essentially, it reverses the process of differentiation.
In our exercise, we find the antiderivative of \( f(x) = 4 - x^2 \) to calculate the definite integral. The steps include:

• Recognizing basic antiderivative rules: For \( 4 \), the antiderivative is \( 4x \); and for \( -x^2 \), the antiderivative is \( -\frac{x^3}{3} \).
• Combining these results, we get: \( \int (4 - x^2) \, dx = 4x - \frac{x^3}{3} + C \).
• Evaluating the antiderivative at the boundaries of the integral, which in this example are \( x = -2 \) and \( x = 2 \).

This computation provides the foundational result needed to perform the definite integration process and subsequently find averages.

Calculus Concepts

The problem of finding the average value of a function over a range involves several key calculus concepts. These include integration, average value theory, and total accumulation.
When discussing average value, remember that it is calculated using:
\[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] where \( [a, b] \) is the interval in question. This formula gives a way to "average out" the total accumulation described by the integral across the interval's length.
To demystify this:

• The integral \( \int_{a}^{b} f(x) \, dx \) represents the total net accumulation of the function \( f(x) \) over \( [a, b] \).
• Dividing this result by the interval’s width \( b-a \) computes its average, as it spreads the total effect evenly across the range.

This is a beautiful aspect of calculus, marrying local rates of change (from derivatives) with macro-level accumulations (through integrals) to provide clear insights into the behavior of functions over intervals.