Problem 17 - Calculus and its applications

Question

Find the average value of \(f(x, y)=2 x-y,\) where \(0 \leq x \leq 2\) and \(2 \leq y \leq 3\).

Step-by-Step Solution

Answer

The average value of the function is \( -\frac{1}{2} \).

Step 1: Understand the Domain

We need to find the average of the function \( f(x, y) = 2x - y \) over a rectangular domain in the plane where \( 0 \leq x \leq 2 \) and \( 2 \leq y \leq 3 \).

Step 2: Define the Average Value Formula

The formula for the average value of a function \( f(x, y) \) over a region \( R \) is given by: \[\text{Average value of } f = \frac{1}{\text{Area of } R} \int \int_R f(x, y) \, dA\] In this case, \( dA = dx \, dy \) and the area of \( R \) is the product of the side lengths: \( 2 - 0 = 2 \) and \( 3 - 2 = 1 \), so \( \text{Area of } R = 2 \times 1 = 2 \).

Step 3: Set Up the Double Integral

The double integral over \( R \) is set up as: \[\int_{0}^{2} \int_{2}^{3} (2x - y) \, dy \, dx\]

Step 4: Integrate with Respect to y

Compute the integral with respect to \( y \): \[\int_{2}^{3} (2x - y) \, dy = \left[ 2xy - \frac{y^2}{2} \right]_2^3 = \left(2x \cdot 3 - \frac{3^2}{2}\right) - \left(2x \cdot 2 - \frac{2^2}{2}\right)\]This simplifies to: \[6x - \frac{9}{2} - (4x - 2) = 2x - \frac{5}{2}\]

Step 5: Integrate with Respect to x

Now, integrate the result from Step 4 with respect to \( x \):\[\int_{0}^{2} \left( 2x - \frac{5}{2} \right) \, dx = \left[ x^2 - \frac{5}{2}x \right]_0^2 = (2^2 - \frac{5}{2}\cdot2) - (0^2 - \frac{5}{2}\cdot0)\]This simplifies to:\[4 - 5 = -1\]

Step 6: Find the Average Value

Use the average value formula:\[\text{Average value} = \frac{-1}{2}\]Thus, the average value of the function \( f(x, y) \) over the region \( R \) is \(-\frac{1}{2}\).

Key Concepts

Average Value of Function

The average value of a function provides a way to summarize the behavior of the function over a particular region. For functions of one variable, this gives us an idea of the function's "typical" value within an interval. When dealing with functions of two variables, represented as \( f(x, y) \), like in our exercise, the concept extends over a two-dimensional region.

To find the average value of the function \( f(x, y) = 2x - y \) over a region, you use the formula: \[ \text{Average value of } f = \frac{1}{\text{Area of } R} \int \int_R f(x, y) \, dA \]This formula divides the integral over the region by the area of the region, \( R \). This gives the average value of \( f(x, y) \) as if its summed effect was distributed evenly across \( R \).

Applying this to our function and region, the formula tells us that while \( f(x, y) = 2x - y \) varies within the region defined by the domains of \( x \) and \( y \), the average behavior of the function across the entire domain results in an average value.

Rectangular Domain

The rectangular domain is a specific type of region in the plane where variables range between specified limits. In our exercise, the region \( R \) is defined by the limits \( 0 \leq x \leq 2 \) and \( 2 \leq y \leq 3 \). This creates a rectangle on the \( xy \)-plane.

A key characteristic of a rectangular domain is its straightforward shape, which simplifies the calculation of integral bounds. Here, the width along the \( x \)-axis is \( 2 - 0 = 2 \) and the height along the \( y \)-axis is \( 3 - 2 = 1 \). This makes calculating the area of \( R \) easy: simply multiply the width by the height, yielding an area of \( 2 \times 1 = 2 \).

This rectangular nature allows for easy setup of the double integral needed to compute the average value. The limits of integration follow directly from the limits of \( x \) and \( y \), making this domain ideal for problems involving double integration.

Integration

Integration is a fundamental concept in calculus used to accumulate quantities, often areas under curves. When dealing with functions of two variables, integration becomes two-dimensional, and this is referred to as double integration.

In a double integral, like the one in our problem, you integrate a function over a specified region. Here's the setup for our exercise: \[ \int_{0}^{2} \int_{2}^{3} (2x - y) \, dy \, dx \] This expression indicates you first integrate \( (2x - y) \) with respect to \( y \), holding \( x \) constant, and then integrate the resulting expression with respect to \( x \).

- **First Integral**: - Here, we integrate \( 2x - y \) with respect to \( y \) from \( 2 \) to \( 3 \). This simplifies to a new expression in terms of \( x \).- **Second Integral**: - Next, we integrate the result from the first integration with respect to \( x \) from \( 0 \) to \( 2 \).

This double integration yields a single number which, divided by the area of the region, gives the aimed average value of the function over that region. In our example, this concluded with an average value of \(-\frac{1}{2}\), showing how the output of double integration coordinates with finding average values.