Question

Find the average value of \(f(x, y)\) over the region \(R\). $$ \begin{aligned} &f(x, y)=e^{x+y}\\\ &R: \text { triangle with vertices }(0,0),(0,1),(1,1) \end{aligned} $$

Short answer

The average value of \(f(x,y) = e^{x+y}\) over the given triangle is \( \frac{e^{2}}{2} - e + \frac{1}{2}\).

Step-by-step solution

Parameters for the Double Integral

The double integral for the average value of a function over a region \R\ in \(x, y\) plane is given by \(\frac{1}{A} \iiiint_{R} f(x, y) dx dy\), where \(A\) is the area of \(R\). In this case an \((x, y)\) region is described by the limits of \(x\) (from 0 to 1) and \(y\) (from 0 to \(x\)). Therefore, the double integral will be between these limits. \(A\) is \(\frac{1}{2}\), since the triangle is right isosceles, with base and height of length 1.

Calculate the Double Integral

Calculate the double integral, and simplify the terms. Now set up the integral: \(\frac{1}{A} \iiiint_{0}^{1} \iiiint_{0}^{x} e^{x+y} dy dx\). The inner integral with respect to \(y\) gives \(e^{x+y}|_{0}^{x}\). Simplifying it gives \(e^{2x}-e^{x}\). The outer integral becomes \(\frac{1}{1/2} \iiiint_{0}^{1} (e^{2x}-e^{x}) dx\).

Solve the Integral

Now, solve the integral: You get \(\frac{1}{2} e^{2x} - e^{x}\Big|_0^1\). After evaluation, you will get \( \frac{1}{2}e^{2} - e - \frac{1}{2} + 1 \) which simplifies to \( \frac{e^{2}}{2} - e + \frac{1}{2}\).

Key concepts

Double Integral

The double integral is a powerful tool in calculus that allows us to calculate the volume under a surface over a certain region. In simple terms, it's like adding up the areas of infinitesimally thin slices of the region to get the total volume. For functions of two variables, such as \( f(x, y) \), the double integral is denoted as \( \iint_{R} f(x, y) \, dx \, dy \), where \( R \) specifies the region over which we are integrating.

When we're given a region like the triangular area described in our exercise, we have to carefully set the limits of integration. These limits define the boundaries of the integration for each variable, and they can sometimes be functions themselves, such as when the upper limit of \( y \) is dependent on the value of \( x \), like in the exercise. Knowing how to correctly set these limits is crucial as it impacts the final result of the integral.

Calculating the double integral involves first integrating with respect to one variable while treating the other as a constant, and then integrating the resulting expression with respect to the second variable, in this case, integrating first with respect to \( y \) and then \( x \). This process often requires a good understanding of both the geometry of the region and the function you're working with.

Exponential Functions

Exponential functions, such as \( e^{x} \), are fundamental in mathematics and occur frequently in various fields, including physics, engineering, and finance. These functions are characterized by the fact that the rate of change (or derivative) of the function at any point is directly proportional to the value of the function at that point.

In our exercise, the function under consideration is \( e^{x+y} \). This exponential function grows rapidly as \( x \) and \( y \) increase. The base \( e \) is an important mathematical constant, approximately equal to 2.71828, and it is the base rate of growth shared by all continually growing processes.

When integrating exponential functions, we often use properties of exponents to simplify the expressions before finding the antiderivatives. For example, knowing that \( e^{x+y} = e^{x}e^{y} \) can be useful in breaking down the function into more manageable parts for integration.

Integration Limits

Integration limits define the boundaries where the function will be integrated. In the context of double integrals, there are two sets of limits: one for each variable, \( x \) and \( y \). Setting the correct limits of integration is vital to obtain the right answer.

For the average value of a function over a specific region, we calculate the extended double integral over the area \( A \) of the region \( R \). Determining the correct integration limits is where the geometry of the region becomes important. As in the given exercise, if the region is a triangle with vertices at \( (0,0), (0,1), (1,1) \), the limits of \( y \) will range from 0 to \( x \) as \( y \) is bound by the line \( y=x \), and \( x \) will be integrated from 0 to 1.

Remember, for non-rectangular regions, the limits of integration can be functions as seen with the upper limit of \( y \), which changes with \( x \), or constants as seen with \( x \). Always visualize the region or sketch it if needed, to determine how \( x \) and \( y \) are related within the boundaries of the region.