Question

Find the average value of \(f(x, y)\) over the region \(R\) where Average value $$=\frac{1}{A} \int_{R} \int f(x, y) d A$$ and where \(A\) is the area of \(R\). \(f(x, y)=e^{x+y}\) \(R:\) triangle with vertices (0,0),(0,1),(1,1)

Short answer

The average value of the function \(f(x, y)\) over the region \(R\) is \(2e - 3\).

Step-by-step solution

Determine the Area of Region R

The region \(R\) is a right triangle with vertices at (0,0), (0,1), and (1,1). The base and the height of the triangle are both equal to 1, so the area \(A\) can be found by using the formula for the area of a triangle \( A = 0.5 \times base \times height = 0.5 \times 1 \times 1 = 0.5.\)

Setting up the Double Integral

To calculate the average value of the function over the region, we will use a double integral. The limits of integration are defined by the bounding values of \(x\) and \(y\) in the region \(R\). For the triangle given, \(y\) varies from 0 to 1, and \(x\) varies from 0 to \(y\). Therefore, the integral is set up as follows: \[ \frac{1}{A} \int_{0}^{1} \int_{0}^{y} e^{x+y} dx dy \]

Solving the Double Integral

First integrate with respect to \(x\): \[ \frac{1}{A} \int_{0}^{1} [e^{x+y}]_{0}^{y} dy, \] which simplifies to \[ \frac{1}{A} \int_{0}^{1} e^{2y} - e^{y} dy. \] Completing the integration with respect to \(y\) will provide the value of the two variable function \(f(x, y)\) over the region \(R\).

Final Calculation and Simplification

Carrying out the calculation and simplifying gives a final result of: \[ Avg value = \frac{1}{0.5} [0.5e^{2y} - e^{y}]_{0}^{1} = 2e - 3. \]