Question

Find the average value of \(f\) over the region \(R .\) Notice how these functions and regions are related to the iterated integrals given in Exercises \(5-8\). \(f(x, y)=3 x^{2}-y+2 ; \quad R\) is bounded by the lines \(y=0\) \(y=2-x / 2\) and \(x=0\).

Short answer

The average value of the function over the region is approximately 22.67.

Step-by-step solution

Define the Boundaries of Region R

The region \( R \) is bounded by the lines \( y=0 \), \( x=0 \), and \( y=2-\frac{x}{2} \). First, solve the equation \( y = 2 - \frac{x}{2} \) to express \( x \) in terms of \( y \); it gives \( x = 4 - 2y \). This means \( x \) ranges from 0 to \( 4-2y \) for any fixed \( y \) value in the range from 0 to 2.

Set Up the Iterated Integral

The average value of a function over a region \( R \) is given by \( \frac{1}{\text{Area}(R)} \int \int_R f(x,y) \ dA \). First, the area of region \( R \) is calculated with the bounds \( x=0 \), \( y=0 \) to \( y=2 \), and \( y=2-\frac{x}{2} \). Then, set up the iterated integral of \( f(x,y) = 3x^2 - y + 2 \) over \( R \): \[ \int_{y=0}^{y=2} \int_{x=0}^{x=4-2y} (3x^2 - y + 2) \ dx \, dy \].

Evaluate the Inner Integral

The inner integral is \( \int_{x=0}^{x=4-2y} (3x^2 - y + 2) \ dx \). First, integrate with respect to \( x \), treating \( y \) as a constant: \[ \int (3x^2 - y + 2) \, dx = \left[ x^3 - yx + 2x \right]_{0}^{4-2y} \]. Compute this: \[ ((4-2y)^3 - y(4-2y) + 2(4-2y)) - (0 - 0 + 0) \].

Simplify the Result of Inner Integral

Calculate the value of the inner integral: \[ (64 - 48y + 12y^2) - 4y + 2y^2 + 8 - 4y \]. Simplifying gives: \[ 64 - 56y + 10y^2 \].

Evaluate the Outer Integral

Now integrate the result from Step 4 with respect to \( y \): \[ \int_{y=0}^{y=2} (64 - 56y + 10y^2) \, dy \]. The integration yields: \[ 64y - 28y^2 + \frac{10}{3}y^3 \] from \( 0 \) to \( 2 \). Compute the definite integral: \[ (128 - 112 + \frac{80}{3}) - (0 - 0 + 0) \].

Calculate the Area of Region R

To find the average value, first compute the area of region \( R \). The region is a triangle with vertices at \((0, 0)\), \((0, 2)\), and \((4, 0)\). The base is 4 units and height is 2 units, so the area is: \[ \frac{1}{2} \times 4 \times 2 = 4 \text{ square units}. \]

Calculate the Average Value

Finally, use the formula for the average value: \( \frac{1}{\text{Area}(R)} \int \int_R f(x,y) \, dA = \frac{1}{4} \times (128 - 112 + \frac{80}{3}) \). Simplifying: \[ \frac{1}{4} \times \frac{272}{3} = \frac{68}{3} \approx 22.67. \]

Key concepts

Iterated Integrals

Iterated integrals help to evaluate the integral of a function over a two-dimensional region by breaking the computation into successive integrations. In the context of finding the average value of a function over a region, iterated integrals are particularly useful.

This process involves:

• First, choosing an order for the integration, usually determined by the problem's setup.
• Integrating the function with respect to one variable while keeping the other variable constant (this is the inner integral).
• The result of the inner integral is then integrated with respect to the other variable (this is the outer integral).

For example, in our solution, after establishing the region of integration, we first integrate the function \(3x^2-y+2\) with respect to \(x\), keeping \(y\) fixed. After this step, we proceed to integrate the resultant expression with respect to \(y\).

Iterated integrals simplify computations by focusing on one dimension at a time, making it easier to handle complex regions and functions. This method is essential in calculating the average value over a specific area.

Region of Integration

The region of integration determines where the function will be evaluated and integrated. Considering geometric features, it provides boundaries that guide the integration limits.

In the given solution, the region \( R \) is defined by the lines \( y=0 \), \( x=0 \), and \( y=2-\frac{x}{2} \).

• These boundaries outline a triangular region with vertices at points \((0,0)\), \((0,2)\), and \((4,0)\).
• Defining the region's shape and size is crucial to setting correct limits for our integrals, ensuring that the integration process captures the full, intended area.

Visualizing this region makes it easier to understand how the function behaves over it and aids significantly in deciding the most efficient way to integrate. The computation of the area, like the triangular region's area in this example, also plays a vital role when calculating the average value of the function over this region.

Integral Calculus

Integral calculus plays a vital role in finding the average value of a function over a defined region. It involves integrating a function to determine areas under curves, which are key elements in solving problems in calculus and beyond.

Here, integral calculus was utilized to:

• First, set up and evaluate the iterated integral of the function \( f(x, y) = 3x^2 - y + 2 \) over the region \( R \).
• Then, determine the total area of the region, calculated using basic geometric principles for triangles.
• Finally, calculate the average value by dividing the integral result by the area of \( R \).

In this exercise, integral calculus complements our understanding and application of iterated integrals and region of integration, leading to the solution. The proper use of these calculus tools allows us to understand not only the behavior of complex functions over various regions but also their overall contribution over those regions. This holistic approach is a cornerstone of integral calculus in mathematics.