Question

Compute the average value of the following functions over the region \(R\). $$f(x, y)=e^{-y} ; R=\\{(x, y): 0 \leq x \leq 6,0 \leq y \leq \ln 2\\}$$

Short answer

Answer: The average value of the function $f(x, y)$ over the region $R$ is $\frac{1}{2 \ln 2}$.

Step-by-step solution

Find the area of the region R

The region R is defined by \(0 \leq x \leq 6\) and \(0 \leq y \leq \ln 2\). It's a rectangle with length 6 and height \(\ln 2\). The area of R is given by: $$ \text{Area}(R) = 6 (\ln 2) $$

Evaluate the double integral

Now let's compute the double integral of \(f(x, y)\) over the region R: $$ \iint_R f(x, y) dA = \int_{0}^{6} \int_{0}^{\ln 2} e^{-y} dy dx $$ First, we'll integrate with respect to y: $$ \int_{0}^{\ln 2} e^{-y} dy = [-e^{-y}]_{0}^{\ln 2} = (-e^{-\ln 2}) + e^0 = (-\frac{1}{2}) + 1 = \frac{1}{2} $$ Now, integrate with respect to x: $$ \int_{0}^{6} \frac{1}{2} dx = \frac{1}{2} [x]_{0}^{6} = \frac{1}{2} (6) = 3 $$

Calculate the average value of the function

Now, we can compute the average value of the function as: $$ \text{Average value} = \frac{1}{\text{Area}(R)}\iint_R f(x, y)\,dA = \frac{1}{6 (\ln 2)}\times 3 = \frac{1}{2 \ln 2} $$ So, the average value of the function \(f(x, y)\) over the region R is \(\frac{1}{2 \ln 2}\).

Key concepts

Average Value of Functions

Calculating the average value of a function over a region allows us to understand how a function behaves on average across that area.
In our example, the average value of the function \( f(x, y) = e^{-y} \) over the specified region \( R \) can be found using a double integral.
The general formula used to discover the average value of a function over a region \( R \) takes the form:

• \[\text{Average value} = \frac{1}{\text{Area of } R}\iint_R f(x, y)\,dA\]

The process involves two main steps:

• Calculating the total sum of the function values over the region by using the double integral \( \iint_R f(x, y)\,dA \).
• Dividing this sum by the area of the region, which standardizes it to find the average.
  This gives us a single number representing the typical value of the function over the region.

Regions of Integration

Regions of integration determine where and over what boundaries we compute our integrals.
They specify the range of values for our variables \( x \) and \( y \). In our example, the region \( R \) was clearly defined as a rectangle:

• \(0 \leq x \leq 6\)
• \(0 \leq y \leq \ln 2\)

This rectangular region provides limits of integration.

• For \( x \), the range is from 0 to 6.
• For \( y \), it's from 0 to \( \ln 2 \).

The area of this rectangle, which is crucial in finding the average value of the function, is simply the product of its width and height:

• Area \( R = 6 \cdot \ln 2 \)

These limits ensure that our integration precisely covers the intended region.

Integration Techniques

Double integration involves integrating a function of two variables over a specified region.
In our problem, we evaluate the integral of \( f(x, y) = e^{-y} \) over the rectangle \( R \).
The integration occurs in two separate stages:

• Inner integral: First, we integrate with respect to \( y \).
  This focuses on a vertical slice of the region, holding \( x \) constant and resolving the integral:\[\int_{0}^{\ln 2} e^{-y} \,dy = \left[-e^{-y}\right]_{0}^{\ln 2} = \frac{1}{2}\]
• Outer integral: Next, we integrate the result with respect to \( x \).
  This step involves summing all these vertical slices across the horizontal span:\[\int_{0}^{6} \frac{1}{2} \,dx = \frac{1}{2} \left[ x \right]_{0}^{6} = 3\]

Integrating in this step-by-step way simplifies the process and ensures accuracy.