The given problem asks for the evaluation of the iterated integral:

$\int_0^{\ln 5} \int_1^{\ln 2} e^{2x + 3y} \, dy \, dx.$

Let’s solve this step by step.

Step 1: Separate the exponential term

The expression $e^{2x + 3y}$ can be written as the product of two independent terms: $e^{2x + 3y} = e^{2x} \cdot e^{3y}.$

This allows us to separate the variables when integrating.

Step 2: Inner integral (with respect to $y$)

We evaluate: $\int_1^{\ln 2} e^{3y} \, dy.$

The integral of $e^{3y}$ is: $\frac{e^{3y}}{3}.$

Apply the limits $y = 1$ to $y = \ln 2$: $\left[ \frac{e^{3y}}{3} \right]_1^{\ln 2} = \frac{e^{3 \ln 2}}{3} - \frac{e^{3 \cdot 1}}{3}.$

Since $e^{3 \ln 2} = (e^{\ln 2})^3 = 2^3 = 8$, this becomes: $\frac{8}{3} - \frac{e^3}{3} = \frac{8 - e^3}{3}.$

Step 3: Outer integral (with respect to $x$)

The outer integral now becomes: $\int_0^{\ln 5} e^{2x} \cdot \frac{8 - e^3}{3} \, dx.$

Since $\frac{8 - e^3}{3}$ is a constant with respect to $x$, we factor it out: $\frac{8 - e^3}{3} \int_0^{\ln 5} e^{2x} \, dx.$

Step 4: Solve the remaining integral

The integral of $e^{2x}$ is: $\frac{e^{2x}}{2}.$

Apply the limits $x = 0$ to $x = \ln 5$: $\left[ \frac{e^{2x}}{2} \right]_0^{\ln 5} = \frac{e^{2 \ln 5}}{2} - \frac{e^{2 \cdot 0}}{2}.$

Since $e^{2 \ln 5} = (e^{\ln 5})^2 = 5^2 = 25$ and $e^0 = 1$, this becomes: $\frac{25}{2} - \frac{1}{2} = \frac{24}{2} = 12.$

Step 5: Combine results

The final result is: $\frac{8 - e^3}{3} \cdot 12 = 4(8 - e^3).$

Final Answer:

The correct choice is: $\textbf{a. } 4(8 - e^3).$

Let me know if you'd like a more detailed explanation!

Related Questions:

1. How does separation of exponential terms work in integrals?
2. Can this problem be solved with reversed integration limits?
3. Why does the constant factor $\frac{8 - e^3}{3}$ simplify the problem?
4. What is the significance of using iterated integrals instead of double integrals?
5. How would the answer change if $e^{2x + 3y}$ was replaced by $e^{2x - 3y}$?

Tip:

Always check whether you can separate variables in an exponential function—it often simplifies multivariable integrals significantly!