Problem 27 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R} x y d A ; R\) is bounded by \(x=0, y=2 x+1,\) and \(y=-2 x+5\).

Step-by-Step Solution

Answer

Question: Evaluate the double integral of the function \(f(x, y) = xy\) over the region R bounded by the lines \(x=0\), \(y=2x+1\), and \(y=-2x+5\). Answer: The value of the double integral is \(\frac{-5}{6}\).

Step 1: Find the Intersection Points

We need to find the intersection points of the given lines: 1. Intersection of \(x=0\) and \(y=2x+1\) $$y = 2(0) + 1 \Rightarrow y = 1$$ The intersection point is \((0, 1)\). 2. Intersection of \(x=0\) and \(y=-2x+5\) $$y = -2(0) + 5 \Rightarrow y = 5$$ The intersection point is \((0, 5)\). 3. Intersection of \(y=2x+1\) and \(y=-2x+5\) Set the equations equal to each other and solve for x: $$2x + 1 = -2x + 5$$ $$4x = 4$$ $$x = 1$$ Now, substitute x into one of the equations to find y: $$y = 2(1) + 1 \Rightarrow y = 3$$ The intersection point is \((1, 3)\). Now, we have the triangular region R determined by the vertices \((0,1)\), \((0,5)\), and \((1,3)\).

Step 2: Set Up the Integrals

Now, we need to set up the integral with the correct bounds. We'll integrate with respect to \(y\) first as it would lead to a simpler integral: $$\iint_{R} xy dA = \int_{x=0}^{x=1} \int_{y=2x+1}^{y=-2x+5} xy dy dx$$

Step 3: Evaluate the Integran[\(\LaTeX\)][Input_in-Math%20Mode]->l

First, we'll evaluate the inner integral: $$\int_{y=2x+1}^{y=-2x+5} xy dy = \frac{1}{2}x y^2 \Big|_{y=2x+1}^{y=-2x+5}$$ $$=\frac{1}{2}x((-2x+5)^2-(2x+1)^2)$$ Now, we'll evaluate the outer integral: $$\int_{x=0}^{x=1} \frac{1}{2}x((-2x+5)^2-(2x+1)^2) dx = \frac{1}{2} \int_{x=0}^{x=1} x(4x^2-28x+24) dx$$ We can use the Power Rule and simplify the integral: $$\frac{1}{2} \int_{x=0}^{x=1} (4x^3 - 28x^2 + 24x) dx$$ $$=\frac{1}{2} \left[\frac{1}{4}(4x^4) - \frac{28}{3}x^3 + 12x^2\right]_{x=0}^{x=1}$$

Step 4: Calculate the Final Answer

Now, we'll evaluate the integral at the bounds:$$\frac{1}{2} \left[\frac{1}{4}(4\cdot1^4) - \frac{28}{3}\cdot1^3 + 12\cdot1^2\right] - \frac{1}{2}\left[\frac{1}{4}(4\cdot0^4) - \frac{28}{3}\cdot0^3 + 12\cdot0^2\right]$$ $$=\frac{1}{2} \left(\frac{1}{4}(4) - \frac{28}{3} + 12\right)$$ $$=\frac{1}{2} \left(1 - \frac{28}{3} + 12\right)$$ $$=\frac{1}{2} \left(\frac{-5}{3}\right)$$ $$=\frac{-5}{6}$$ The value of the double integral is \(\frac{-5}{6}\).

Key Concepts

Integration Techniques

Understanding the methods for integrating functions is essential for evaluating double integrals. In our exercise, integration techniques allow us to compute the area beneath curves by partitioning the region into infinitesimal elements and summing their contributions. For example, the Power Rule, an essential integration technique, was used to integrate polynomial functions in the problem solution. It states that the integral of a function in the form of \(x^n\) is \(\dfrac{x^{n+1}}{n+1}\) for \(n\eq-1\). This rule simplifies the process by providing a straightforward formula for finding antiderivatives of polynomials, as shown when we computed the outer integral in the solution. To achieve mastery in evaluating double integrals, one must practice various integration techniques, which include but are not limited to substitution, integration by parts, and trigonometric integration. These techniques become tools that enable students to handle integrals of varying complexity, ensuring they can tailor their approach to the specific problem at hand.

Utilizing visual aids, such as a graph of the region of integration, can considerably aid in conceptualizing the problem and its solution. A clear, well-labeled sketch illustrates the region being integrated over, the direction of integration, and how the limits of integration are determined.

Area Bounded by Curves

When solving double integrals, the concept of finding the area bounded by curves is crucial. In the exercise, we determine the area in a two-dimensional plane enclosed by the curves described by the equations \(x=0\), \(y=2x+1\), and \(y=-2x+5\). The region of integration, denoted by \(R\), embodies the space within these boundaries where the function is integrated.

The order in which we integrate with respect to \(x\) and \(y\), known as the order of integration, can be chosen based on convenience and the shape of the region. In general, if the region is simpler to describe with one variable held constant, that variable should be the limit of the outer integral. In our example, integrating with respect to \(y\) first and then \(x\) allows us to easily express the bounds in terms of constants and the variable \(x\). The computation of the area under the curves then involves the application of the double integral over these limits. The integral's value numerically equals the area, which can be interpreted as the accumulation of all the 'tiny pieces' of area \(dA\) over the entire region \(R\).

Intersection Points

Determining intersection points is often a preliminary step in evaluating double integrals, especially when the region of integration is bounded by curves. These points of intersection define the limits of integration. In the given exercise, we found the intersection points of the curves to identify the vertices of the triangular region over which we integrated. Calculating intersection points requires solving a system of equations, as demonstrated in the solution where two lines \(y=2x+1\) and \(y=-2x+5\) were set equal to find their point of crossing.

For the correct evaluation of the double integral, it is imperative to understand the geometric significance of these points. They provide the boundaries that contain the area we are calculating. After determining these points, they must be correctly placed in the integral's limits to ensure the function is integrated over the exact region of interest. In practice, errors in identifying intersection points can lead to incorrect limits of integration and thus an incorrect evaluation of the integral.