Problem 18 - Calculus and its applications

Question

Find the average value of \(g(x, y)=4-x-y,\) where \(-1 \leq x \leq 1\) and \(-2 \leq y \leq 3\).

Step-by-Step Solution

Answer

The average value is 2.3.

Step 1: Understand the Variables and Limits

The function given is \(g(x, y) = 4 - x - y\). It is defined over the rectangle in the xy-plane where \(-1 \leq x \leq 1\) and \(-2 \leq y \leq 3\). We need to find the average value of this function over the given region.

Step 2: Find the Area of Integration Region

The region of integration is a rectangle. The length along the x-axis is \(1 - (-1) = 2\), and the length along the y-axis is \(3 - (-2) = 5\). Therefore, the area of the region, \(A\), is \(2 \times 5 = 10\).

Step 3: Set Up the Double Integral for the Average Value

The average value \( \bar{g} \) of the function \(g(x, y)\) over the region can be determined using the formula \[ \bar{g} = \frac{1}{A} \int_{-1}^{1} \int_{-2}^{3} (4-x-y) \, dy \, dx \] where \(A = 10\).

Step 4: Solve the Inner Integral

First, integrate the function \(4-x-y\) with respect to \(y\):\[ \int_{-2}^{3} (4 - x - y) \, dy = \left[ (4-x)y - \frac{y^2}{2} \right]_{-2}^{3}. \]Substitute the limits for \(y\):\[ [(4-x) \cdot 3 - \frac{3^2}{2}] - [(4-x) \cdot (-2) - \frac{(-2)^2}{2}]. \]This simplifies to: \[ (12 - 3x - 4.5) + (8 - 2x - 2) = 13 - 5x. \]

Step 5: Solve the Outer Integral

Now integrate the result from the inner integral with respect to \(x\):\[ \int_{-1}^{1} (13 - 5x) \, dx = \left[ 13x - \frac{5x^2}{2} \right]_{-1}^{1}. \]Substituting the limits for \(x\):\[ (13 \cdot 1 - \frac{5 \cdot 1^2}{2}) - (13 \cdot (-1) - \frac{5 \cdot (-1)^2}{2}). \]Simplifying gives: \[ (13 - 2.5) - (-13 - 2.5) = 23. \]

Step 6: Calculate the Average Value

Finally, use the integral found to compute the average value:\[ \bar{g} = \frac{1}{10} \times 23 = 2.3. \]

Key Concepts

Average Value of Function

The average value of a function describes what the mean value of a function is over a specified region. It gives a sense of what the "central tendency" is for the function within that area. For any function of two variables, like the given function \( g(x, y) = 4 - x - y \), computing the average value means evaluating how the function behaves across all points in the region and then taking the mean of these values.

To calculate the average value of \( g(x, y) \) across a rectangular region, we use the formula:

\[ \bar{g} = \frac{1}{A} \int_{R} g(x,y) \, dA \]

• \( A \) is the area of the region over which we are averaging.
• \( \int_{R} g(x,y) \, dA \) denotes the double integral of the function \( g(x, y) \) over the region \( R \).

The output of this integral is the total "accumulated" value of the function over that area. Dividing by \( A \) then gives us the average value.

Region of Integration

The concept of the region of integration is crucial when dealing with double integrals. It defines the specific area in the xy-plane over which integration will be performed. In our current problem, the region of integration is a rectangle, defined by the constraints:

• \(-1 \leq x \leq 1\)
• \(-2 \leq y \leq 3\)

These limits tell us that the region is a rectangle stretching from \(-1\) to \(1\) along the x-axis and from \(-2\) to \(3\) on the y-axis. The total area of this region is the product of its dimensions, which is \(2\) units along the x-axis and \(5\) units along the y-axis, giving \(A = 10\) square units.

This specific rectangle represents where the function \( g(x, y) = 4 - x - y \) is sampled to compute the average.

Inner Integral

The first step in evaluating a double integral is solving the inner integral. The inner integral considers integration along one variable while keeping the other constant. For the problem at hand, the inner integral is with respect to \( y \) and is represented as:

\[ \int_{-2}^{3} (4 - x - y) \, dy \]

This calculation allows us to summarize how the function behaves over the entire y-range for each fixed value of x. Performing this integration gives us an expression that we can then integrate with respect to x. The result from evaluating this integral was \(13 - 5x\).

It is important to perform the step-by-step integration from limits \(-2\) to \(3\) for \(y\) to ensure all possible contributions across this dimension are counted before moving to the outer integral.

Outer Integral

With the inner integral resolved, the next step is to compute the outer integral. This integral handles the remaining variable—in this case, \( x \). Here's the outer integral based on the result of the inner integration:

\[ \int_{-1}^{1} (13 - 5x) \, dx \]

By integrating this expression with respect to \(x\), we account for the combined effect of \( g(x, y) \) across the region. The evaluation of this outer integral yields a value of \( 23 \).

It's the final step to summarize how the function interacts as it narrows down the rectangle's width on the x-dimension. After this, the last step is dividing by the area \( A \) to compute the average value which, in this problem, results in \( \bar{g} = 2.3 \).