Problem 2 - Calculus an Applied Approach

Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$

Step-by-Step Solution

Answer

The region of integration is a rectangle in the xy-plane with vertices (0,0), (1,0), (0,3), and (1,3). The value of the given double integral over this region is 18.

Step 1: Understand the region of integration

In this double integral the limits of \(y\) are 0 to 1, and for \(x\) are 0 to 3. Hence, the region of integration is a rectangle in the xy-plane with vertices (0,0), (1,0), (0,3), and (1,3).

Step 2: Perform inner integral

Firstly, perform the integral \(\int_{0}^{1}(2 x+6 y) dy\). Here, \(x\) is considered as a constant, and the integral is with respect to \(y\). The antiderivative of \(2x\) w.r.t \(y\) is \(2xy\), and the antiderivative of \(6y\) w.r.t \(y\) is \(3y^{2}\). So,\[\int_{0}^{1}(2 x+6 y) dy = [2xy + 3y^{2}]_{0}^{1} = 2x(1) + 3(1)^{2} - 2x(0) + 3(0)^{2}= 2x + 3\]

Step 3: Perform outer integral

Secondly, perform the integral \(\int_{0}^{3}(2x + 3) dx\), with respect to \(x\). The antiderivative of \(2x\) w.r.t \(x\) is \(x^{2}\), and the antiderivative of \(3\) w.r.t \(x\) is \(3x\). So,\[\int_{0}^{3}(2x + 3) dx = [x^{2} + 3x]_{0}^{3} = 3^{2} + 3(3) - (0)^{2} + 3(0) = 18\]

Key Concepts

Region of Integration

The region of integration is the part of the xy-plane that is defined by the limits of the integrals in a double integral.In our example, we have the double integral:\[\int_{0}^{3} \int_{0}^{1}(2 x+6 y) dy \, dx\]The limits of integration for \(y\) are from 0 to 1, and for \(x\) are from 0 to 3.

This means that our region is a simple rectangle in the xy-plane.It has these four vertices: (0,0), (1,0), (0,3), and (1,3). - Think of this rectangle as the zone over which we compute the integrals.- The bounds correspond to how far we "stretch" or "slice" the area on the x and y axes. Visualizing this rectangle can help you better understand where the integral takes place.

Antiderivative

When we compute a definite integral, we evaluate its antiderivative over the bounds of integration. For the inner integral of \( \int (2x + 6y) \, dy \), we need the antiderivative with respect to \(y\):- For \(2x\), since \(x\) is a constant in this context, the antiderivative is simply \(2xy\).- For \(6y\), the antiderivative is \(3y^2\), because the antiderivative of \(y\) is \(\frac{y^2}{2}\).So, the antiderivative of \(2x + 6y\) with respect to \(y\) is \(2xy + 3y^2\).

For the outer integral of \( \int (2x + 3) \, dx \):- The antiderivative for \(2x\) is \(x^2\).- Constant \(3\) integrates to \(3x\).Finding antiderivatives allows us to solve the integral by plugging in the boundary values.

Inner Integral

The inner integral in a double integral is the first integral you solve.
In our example \( \int_{0}^{1} (2x + 6y) \, dy \), this is performed with respect to \(y\) first.- Treat \(x\) as a constant while integrating over \(y\).- We found that the antiderivative is \(2xy + 3y^2\).Once we find the general antiderivative, we evaluate it from 0 to 1:\[[2xy + 3y^2]_{0}^{1} = 2x(1) + 3(1)^2 - 2x(0) - 3(0)^2 = 2x + 3\]This is the solution for the inner integral, which becomes the input for the outer integral.

Outer Integral

The outer integral is performed after completing the inner integral.It utilizes the results from the inner integral solution as its new integrand. In our example, the outer integral is:\[\int_{0}^{3} (2x + 3) \, dx\]We integrate this with respect to \(x\). - From the inner integral, we got \(2x + 3\) as our new function to integrate.- We compute the antiderivative of \(2x\) which is \(x^2\), and \(3\) which becomes \(3x\).We then evaluate this antiderivative from 0 to 3:\[[x^2 + 3x]_{0}^{3} = 3^2 + 3 \times 3 - (0)^2 - 3 \times 0 = 18\]The result of the outer integral, 18, is the final answer to the original double integral problem.