Problem 1
Question
In Exercises \(1-8,\) sketch the region of integration and evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{1}(3 x+4 y) d y d x $$
Step-by-Step Solution
Verified Answer
The value of the double integral is 10.
1Step 1: Sketch the region of integration
Here, \(x\) and \(y\) both vary from 0 to 2 and 1 respectively. This creates a rectangle in the \(xy\)-plane with vertices at (0,0), (2,0), (2,1), and (0,1).
2Step 2: Evaluate the Inner Integral (Integrate with respect to \(y\))
Treat \(x\) as a constant, and integrate the function \(3x + 4y\) with respect to \(y\). The definite integral is given by \(\int_{0}^{1} (3x+4y) dy = 3x*y + 2*y^2 |_{0}^{1} = 3x + 2\).
3Step 3: Evaluate the Outer Integral (Integrate with respect to \(x\))
Now, integrate the result of step 2 with respect to \(x\). That is, evaluate \(\int_{0}^{2} (3x + 2) dx = \frac{3}{2}*x^2 + 2x |_{0}^{2} = 6 + 4 =10\).
Key Concepts
Integration TechniquesDefinite IntegralRegion of Integration Sketch
Integration Techniques
Integration is a fundamental tool in calculus that is used to find the area under a curve, among other applications. In the case of a double integral, such as
\[\int_{0}^{2} \int_{0}^{1}(3 x+4 y) d y d x \],
multiple integration techniques can be applied to evaluate the integral. One common technique is to first treat any variables not related to the integration variable as constants, as was demonstrated in the exercise solution here. This method tackles the integration process in layers, starting from the innermost integral and working outward.
In this example, during the evaluation of the inner integral with respect to \(y\), \(x\) is considered as a constant, thereby simplifying the process. Then, the integrand
\(3x + 4y\)
is integrated with respect to \(y\), yielding a function of \(x\) alone. This process is repeated for the outer integral, integrating the resulting expression with respect to \(x\). This layer-by-layer approach is particularly useful for functions that are separable in their variables, such as polynomials, exponentials, and trigonometric functions.
\[\int_{0}^{2} \int_{0}^{1}(3 x+4 y) d y d x \],
multiple integration techniques can be applied to evaluate the integral. One common technique is to first treat any variables not related to the integration variable as constants, as was demonstrated in the exercise solution here. This method tackles the integration process in layers, starting from the innermost integral and working outward.
In this example, during the evaluation of the inner integral with respect to \(y\), \(x\) is considered as a constant, thereby simplifying the process. Then, the integrand
\(3x + 4y\)
is integrated with respect to \(y\), yielding a function of \(x\) alone. This process is repeated for the outer integral, integrating the resulting expression with respect to \(x\). This layer-by-layer approach is particularly useful for functions that are separable in their variables, such as polynomials, exponentials, and trigonometric functions.
Definite Integral
A definite integral has specific limits or 'bounds' of integration, and it represents the accumulation of quantities like area or volume, depending on the dimensionality. In the given exercise, we encounter a double definite integral with bounds from \(0\) to \(1\) for \(y\) and from \(0\) to \(2\) for \(x\).
When evaluated, a definite integral produces a numerical value, as opposed to an indefinite integral, which results in a function plus an arbitrary constant. For the inner integral, we integrated \(3x+4y\) with respect to \(y\) and applied the bounds \(0\) to \(1\). Then for the outer integral, we integrated with respect to \(x\), applying the bounds \(0\) to \(2\), which ultimately yielded the numerical solution \(10\).
When evaluated, a definite integral produces a numerical value, as opposed to an indefinite integral, which results in a function plus an arbitrary constant. For the inner integral, we integrated \(3x+4y\) with respect to \(y\) and applied the bounds \(0\) to \(1\). Then for the outer integral, we integrated with respect to \(x\), applying the bounds \(0\) to \(2\), which ultimately yielded the numerical solution \(10\).
- The evaluation of the outer integral is done after substituting the bounds of the inner integral.
- It encapsulates the area of the region described by the bounds of integration.
Region of Integration Sketch
Visualizing the region of integration is a crucial step in evaluating double integrals, as it provides insight into the geometry of the problem. Sketching the region helps to verify the correct application of the limits of integration.
In our example, the region of integration is a rectangle in the \(xy\)-plane defined by the bounds \(0\) ≤ \(x\) ≤ \(2\) and \(0\) ≤ \(y\) ≤ \(1\). The four vertices of this rectangle are \((0,0)\), \((2,0)\), \((2,1)\), and \((0,1)\). It is essential to ensure that the sketch is accurate, as a misunderstanding of the region can lead to errors in the setup and execution of the integral.
In our example, the region of integration is a rectangle in the \(xy\)-plane defined by the bounds \(0\) ≤ \(x\) ≤ \(2\) and \(0\) ≤ \(y\) ≤ \(1\). The four vertices of this rectangle are \((0,0)\), \((2,0)\), \((2,1)\), and \((0,1)\). It is essential to ensure that the sketch is accurate, as a misunderstanding of the region can lead to errors in the setup and execution of the integral.
Benefits of Sketching the Region:
- Serves as a visual aid in understanding the limits of integration.
- Allows for the identification of symmetries and geometrical properties that might simplify the integration process.
- Is helpful for recognizing instances where alternative methods like polar, cylindrical, or spherical coordinates may be beneficial.
Other exercises in this chapter
Problem 1
In Exercises \(1-10,\) evaluate the partial integral. $$ \int_{0}^{x}(2 x-y) d y $$
View solution Problem 1
Find any critical points and relative extrema of the function. $$ f(x, y)=x^{2}-y^{2}+4 x-8 y-11 $$
View solution Problem 1
Find the first partial derivatives with respect to \(x\) and with respect to \(y .\) $$ z=3 x+5 y-1 $$
View solution