Question

Sketch the region of integration and evaluate the following integrals as they are written. $$\int_{0}^{4} \int_{y}^{2 y} x y d x d y$$

Short answer

#Answer# The value of the given double integral is 96.

Step-by-step solution

Sketch the region of integration

To sketch the region of integration, we first look at the limits of integration. The outer integral has a limit from 0 to 4, and the inner integral has a limit from y to 2y in terms of x. The region is defined by the lines x=y and x=2y for y between 0 and 4. To draw this region, simply plot the lines x=y and x=2y for y in the range [0, 4], and fill the area between these lines.

Integrate with respect to x

Now we will calculate the inner integral with respect to x: $$\int_{y}^{2 y} x y d x$$ To do this, integrate the function xy with respect to x: $$\left[\frac{1}{2}x^2y\right]_{x=y}^{x=2y}$$ Now, substitute the limits: $$\frac{1}{2}(2y)^2y - \frac{1}{2}(y)^2y = \frac{3}{2}y^3$$

Integrate with respect to y

Next, we will calculate the outer integral with respect to y: $$\int_{0}^{4} \frac{3}{2}y^3 d y$$ To do this, integrate the function \(\frac{3}{2}y^3\) with respect to y: $$\left[\frac{3}{8}y^4\right]_{y=0}^{y=4}$$ Now, substitute the limits: $$\frac{3}{8}(4)^4 - \frac{3}{8}(0)^4 = \frac{3}{8}(256) = 96$$ The value of the given double integral is 96.

Key concepts

Region of Integration

The region of integration is a fundamental aspect in evaluating double integrals. For this exercise, it includes examining the area over which the integration occurs. In our case, the limits tell us the integration region lies between the curves described by the lines \(x = y\) and \(x = 2y\). These lines create boundaries over which the variables are integrated.

The outer integral runs from \(y = 0\) to \(y = 4\), suggesting that \(y\) values span this range. For each specific \(y\), \(x\) ranges between \(y\) and \(2y\).

To visualize this, consider drawing lines on the coordinate plane:

• The line \(x = y\) is a diagonal passing through the origin, inclining at a 45-degree angle.
• The line \(x = 2y\) is steeper, inclined at approximately 63.43 degrees, passing through points like (0,0) and (4,2).

Shade the area between these two lines from \(y = 0\) to \(y = 4\). This shaded area represents the region of integration for our integral.

Integration Limits

Understanding integration limits is crucial for setting up and solving double integrals. Integration limits dictate the bounds of integration for each variable in an iterated integral.

In our integral setup \[\int_{0}^{4} \int_{y}^{2y} x y \, dx \,dy\],

• The inner integral \(\int_{y}^{2y}\) represents \(x\)'s limits. For a specific \(y\), \(x\) spans from \(y\) to \(2y\).
• The outer integral \(\int_{0}^{4}\) comprehends \(y\)'s limits, ranging from 0 to 4.

The inner integration limit critically depends on \(y\). As \(y\) increases, \(x\)'s permissible range also shifts and extends from \(y\) to \(2y\). Therefore, the limits create a varying boundary for \(x\) across the integration, forming a trapezoidal region of integration.

Mastering how to interpret these limits ensures accurate setting of regions, helping in solving integrals effectively.

Iterated Integrals

Iterated integrals are essentially a step-by-step method of evaluating double integrals. This process involves nested integration, where each level of the integral represents a different variable.

In the given exercise, the double integral is represented as \[\int_{0}^{4} \left( \int_{y}^{2y} xy \, dx \right) dy.\] Here:

• The inner integral \(\int_{y}^{2y} xy \, dx\) focuses on \(x\). We treat \(y\) as a constant here to solve it separately.
• The outer integral \(\int_{0}^{4} \, dy\) operates on \(y\). After determining the result of the inner integral, you integrate with respect to \(y\).

Calculating an iterated integral involves resolving the inner integral first and then using this result for the outer integral. This sequential resolution simplifies multiple-variable integrals into manageable, single-variable ones. For our problem, solving these integrals individually leads us to the result: 96.