To solve the given iterated integral, let's analyze the expression in detail.

The iterated integral is:

$\int_{1}^{4} \int_{1}^{2} \left( \frac{x}{y} + \frac{y}{x} \right) \, dx \, dy$

This integral can be solved by following these steps:

Step 1: Separate the Inner Integral

Rewrite the integral by separating the terms inside the integrand:

$\int_{1}^{4} \int_{1}^{2} \left( \frac{x}{y} + \frac{y}{x} \right) \, dx \, dy = \int_{1}^{4} \int_{1}^{2} \frac{x}{y} \, dx \, dy + \int_{1}^{4} \int_{1}^{2} \frac{y}{x} \, dx \, dy$

Step 2: Solve Each Inner Integral Separately

We will solve each part of the integral separately.

Part 1: Compute $\int_{1}^{2} \frac{x}{y} \, dx$

Since $\frac{x}{y}$ does not depend on $y$ (it treats $y$ as a constant with respect to $x$), we can take $\frac{1}{y}$ outside the inner integral:

$\int_{1}^{4} \int_{1}^{2} \frac{x}{y} \, dx \, dy = \int_{1}^{4} \frac{1}{y} \left( \int_{1}^{2} x \, dx \right) dy$

Now, solve the inner integral $\int_{1}^{2} x \, dx$:

$\int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$

Substitute back:

$\int_{1}^{4} \frac{1}{y} \cdot \frac{3}{2} \, dy = \frac{3}{2} \int_{1}^{4} \frac{1}{y} \, dy$

Now, integrate with respect to $y$:

$\frac{3}{2} \int_{1}^{4} \frac{1}{y} \, dy = \frac{3}{2} \left[ \ln y \right]_{1}^{4} = \frac{3}{2} (\ln 4 - \ln 1) = \frac{3}{2} \ln 4$

Since $\ln 1 = 0$, we have:

$\frac{3}{2} \ln 4$

Part 2: Compute $\int_{1}^{2} \frac{y}{x} \, dx$

Similarly, treat $\frac{y}{x}$ as a constant with respect to $y$ and take $y$ outside the inner integral:

$\int_{1}^{4} \int_{1}^{2} \frac{y}{x} \, dx \, dy = \int_{1}^{4} y \left( \int_{1}^{2} \frac{1}{x} \, dx \right) dy$

Now, solve the inner integral $\int_{1}^{2} \frac{1}{x} \, dx$:

$\int_{1}^{2} \frac{1}{x} \, dx = \left[ \ln x \right]_{1}^{2} = \ln 2 - \ln 1 = \ln 2$

Substitute back:

$\int_{1}^{4} y \cdot \ln 2 \, dy = \ln 2 \int_{1}^{4} y \, dy$

Now, integrate with respect to $y$:

$\ln 2 \int_{1}^{4} y \, dy = \ln 2 \left[ \frac{y^2}{2} \right]_{1}^{4} = \ln 2 \left( \frac{4^2}{2} - \frac{1^2}{2} \right)$

Calculate each term:

$= \ln 2 \left( \frac{16}{2} - \frac{1}{2} \right) = \ln 2 \cdot \frac{15}{2} = \frac{15}{2} \ln 2$

Step 3: Combine the Results

Add the results from both parts:

$\int_{1}^{4} \int_{1}^{2} \left( \frac{x}{y} + \frac{y}{x} \right) \, dx \, dy = \frac{3}{2} \ln 4 + \frac{15}{2} \ln 2$

Since $\ln 4 = 2 \ln 2$, substitute $\ln 4$ to simplify further:

$= \frac{3}{2} \cdot 2 \ln 2 + \frac{15}{2} \ln 2 = \frac{6}{2} \ln 2 + \frac{15}{2} \ln 2 = \frac{21}{2} \ln 2$

Final Answer

$\int_{1}^{4} \int_{1}^{2} \left( \frac{x}{y} + \frac{y}{x} \right) \, dx \, dy = \frac{21}{2} \ln 2$

Would you like further details on any specific part?

Related Questions

1. How do you apply the Fubini’s theorem to evaluate double integrals?
2. Can iterated integrals be solved by reversing the order of integration?
3. How does the choice of order in integration affect the complexity of double integrals?
4. What are the conditions for separating integrals when the integrand has multiple terms?
5. How would this integral change if the bounds were reversed?

Tip

When dealing with iterated integrals, check if you can simplify by separating terms or factoring out constants to make the integration manageable.