Problem 26 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Evaluate the given double integral for the specified region \(R\).\(\iint_{R}(x+2 y) d A\), where \(R\) is the triangle with vertices \((0,0),(1,0)\), and \((0,2)\).

Step-by-Step Solution

Answer

The value of the double integral is \( \frac{5}{3} \).

Step 1 - Define the region

The region \(R\) is a triangle with vertices at \((0,0)\), \((1,0)\), and \((0,2)\). Visualize this triangle in the coordinate plane. The equation of the line connecting \((0,2)\) and \((1,0)\) is \(y = 2 - 2x\). Thus, \(R\) can be described as the set of points \((x,y)\) where \(0 \leq x \leq 1\) and \(0 \leq y \leq 2 - 2x\).

Step 2 - Set up the double integral

Set up the double integral with the determined bounds for the region \(R\). The integrand is \(f(x,y) = x + 2y\). Hence, the integral becomes: \[ \iint_{R} (x + 2y) \, dA = \int_{0}^{1} \int_{0}^{2-2x} (x + 2y) \, dy \, dx \]

Step 3 - Integrate with respect to \(y\)

First, integrate \(x + 2y\) with respect to \(y\): \[ \int_{0}^{2-2x} (x + 2y) \, dy \]Using the antiderivative, we get:\[ xy + y^2 \bigg|_{0}^{2-2x} \]Evaluate the bounds:\[ \left[ x(2-2x) + (2-2x)^2 \right] - [x(0) + 0^2] = (2x - 2x^2) + (4 - 8x + 4x^2) \]=\[ 2x - 2x^2 + 4 - 8x + 4x^2 = 2x^2 - 6x + 4\]

Step 4 - Integrate with respect to \(x\)

Now, integrate the resulting expression with respect to \(x\): \[ \int_{0}^{1} (2x^2 - 6x + 4) \, dx \]Using the antiderivative, we get:\[ \frac{2x^3}{3} - 3x^2 + 4x \bigg|_{0}^{1} \]Evaluate the bounds:\[ \left[ \frac{2(1)^3}{3} - 3(1)^2 + 4(1) \right] - \left[ \frac{2(0)^3}{3} - 3(0)^2 + 4(0) \right] = \frac{2}{3} - 3 + 4 = \frac{2}{3} + 1 = \frac{5}{3} \]

Key Concepts

Double Integrals

A double integral is a way to integrate over a two-dimensional region. Think of it as an extension of a single integral, which only integrates along one dimension. It is often used in calculus to find the volume under a surface defined by a function of two variables, like the function in our example: \ f(x, y) = x + 2y. You can represent a double integral as follows: \ \[\iint_{R} f(x,y) \, dA\] \To solve a double integral, you typically break it down into two single integrals. You first integrate with respect to one variable, holding the other constant, and then you integrate the result with respect to the second variable. In our example, the integration is done first with respect to y and then x.

Region of Integration

Defining the region of integration is crucial for solving double integrals. The region tells us the bounds over which we are integrating. In this exercise, the region R is a triangle with vertices at (0,0), (1,0), and (0,2). \br The equation of the line connecting the points (0,2) and (1,0) can be found using the slope-intercept method, which yields y = 2 - 2x.\br Hence, the region R can be described using inequalities: \[ 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 2 - 2x \]These inequalities are used as bounds when setting up the double integral. Understanding the region is critical as it ensures the correct limits for our integration process.

Antiderivative

An antiderivative, or indefinite integral, is a function whose derivative is the original function. It reverses the process of differentiation. When resolving double integrals, finding the antiderivative is necessary to evaluate the integral.\br In our particular step-by-step solution, we handled the integral \(\int_{0}^{2-2x} (x + 2y) \, dy \). We first integrated with respect to y. The antiderivative of x is xy, and the antiderivative of 2y is y^2. So the solution to the integral is:\br \[ xy + y^2 \bigg|_{0}^{2-2x} \]\When evaluating an antiderivative at its bounds, we substitute and subtract: \[[ x(2-2x) + (2-2x)^2 \]]\This step-by-step method showcases the importance of correctly finding and evaluating the antiderivative in solving double integrals.