91Ó°ÊÓ

When dealing with double integrals, the concept of iterated integrals comes into play. This involves breaking down a double integral into separate single integrals, where the integration occurs in one variable at a time. In the given exercise, the integration over the region \(R\) is expressed as an iterated integral: \[ \iint_{R} e^{x+2 y} dA = \int_1^{\ln3} \int_0^{\ln2} e^{x+2 y} dx \, dy \]This setup reflects the order of integration, starting with \(x\) ranging from \(0\) to \(\ln 2\) and then integrating with respect to \(y\) from \(1\) to \(\ln 3\). This method simplifies the process by focusing on one dimension at a time.
To set up an iterated integral correctly, it’s crucial to carefully define the limits of integration, which match the bounds of the region \(R\) over which the integration is performed. This meticulous setup allows you to tackle complex regions in a structured way.

In solving iterated integrals, employing the right integration technique is key. To begin with, integrating with respect to \(x\) involves treating \(y\) as a constant. The function inside the integral, \(e^{x+2y}\), becomes easier to handle due to its exponential form. For exponential functions, integrating involves finding an antiderivative, which is straightforward since the derivative and antiderivative of \(e^x\) is \(e^x\) itself. Once you integrate over \(x\), you are left with a simpler expression in terms of \(y\), particularly \(e^{2y}\) in this case. Requiring similar techniques, you proceed to integrate with respect to \(y\).
When \(y\) is the active variable, integration of \(e^{2y}\) yields \(\frac{1}{2}e^{2y}\), highlighting how each step simplifies down naturally. Understanding these integration techniques simplifies handling double integrals, especially when using substitution or dealing with logarithmic functions.

Evaluating the iterated integral effectively closes the process. After integrating both with respect to \(x\) and \(y\), evaluating involves substituting the limits into the expressions you’ve found. This step transforms the expressions into specific numeric values. In our example, after integrating and substituting limits, you arrive at:\[ \frac{1}{2}(e^{2\ln3} - e^{2}) = \frac{1}{2}(e^{\ln9} - e^{2}) \]This expression simplifies further because \(e^{\ln9}\) equals 9. Here’s where simplification comes into full play as you substitute back any numeric terms involved:\[ \frac{1}{2}(9 - e^2) \]
This final step might require additional arithmetic or algebraic manipulation to get to the simplest form or final numeric result. Each evaluation not only provides the value of the double integral but reinforces understanding by synthesizing different integration rules.