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The problem asks to verify the equality of two iterated double integrals:

\[
\int_1^2 \left( \int_3^4 \left( xy + e^y \right) dx \right) dy = \int_3^4 \left( \int_1^2 \left( xy + e^y \right) dy \right) dx
\]

This is an example of Fubini's Theorem, which allows us to switch the order of integration as long as the function is continuous and the limits are constants.

### Step 1: Evaluate the Left-Hand Side (LHS)

\[
\text{LHS} = \int_1^2 \left( \int_3^4 \left( xy + e^y \right) dx \right) dy
\]

Let's break this down by first evaluating the inner integral with respect to \( x \):

\[
\int_3^4 \left( xy + e^y \right) dx = \int_3^4 xy \, dx + \int_3^4 e^y \, dx
\]

For the first term:
\[
\int_3^4 xy \, dx = y \int_3^4 x \, dx = y \left[ \frac{x^2}{2} \right]_3^4 = y \left( \frac{16}{2} - \frac{9}{2} \right) = y \left( \frac{7}{2} \right)
\]

For the second term, since \( e^y \) is constant with respect to \( x \):
\[
\int_3^4 e^y \, dx = e^y \cdot (4 - 3) = e^y
\]

Thus, the inner integral becomes:
\[
\int_3^4 \left( xy + e^y \right) dx = \frac{7y}{2} + e^y
\]

Now, integrate this with respect to \( y \) from 1 to 2:
\[
\int_1^2 \left( \frac{7y}{2} + e^y \right) dy = \frac{7}{2} \int_1^2 y \, dy + \int_1^2 e^y \, dy
\]

For the first term:
\[
\frac{7}{2} \int_1^2 y \, dy = \frac{7}{2} \left[ \frac{y^2}{2} \right]_1^2 = \frac{7}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{7}{2} \cdot \frac{3}{2} = \frac{21}{4}
\]

For the second term:
\[
\int_1^2 e^y \, dy = e^y \Big|_1^2 = e^2 - e
\]

Thus, the left-hand side evaluates to:
\[
\text{LHS} = \frac{21}{4} + (e^2 - e)
\]

### Step 2: Evaluate the Right-Hand Side (RHS)

\[
\text{RHS} = \int_3^4 \left( \int_1^2 \left( xy + e^y \right) dy \right) dx
\]

Again, break this down by first evaluating the inner integral with respect to \( y \):

\[
\int_1^2 \left( xy + e^y \right) dy = x \int_1^2 y \, dy + \int_1^2 e^y \, dy
\]

For the first term:
\[
x \int_1^2 y \, dy = x \left[ \frac{y^2}{2} \right]_1^2 = x \left( \frac{4}{2} - \frac{1}{2} \right) = x \cdot \frac{3}{2}
\]

For the second term:
\[
\int_1^2 e^y \, dy = e^y \Big|_1^2 = e^2 - e
\]

Thus, the inner integral becomes:
\[
\int_1^2 \left( xy + e^y \right) dy = \frac{3x}{2} + (e^2 - e)
\]

Now, integrate this with respect to \( x \) from 3 to 4:
\[
\int_3^4 \left( \frac{3x}{2} + (e^2 - e) \right) dx = \frac{3}{2} \int_3^4 x \, dx + (e^2 - e) \int_3^4 1 \, dx
\]

For the first term:
\[
\frac{3}{2} \int_3^4 x \, dx = \frac{3}{2} \left[ \frac{x^2}{2} \right]_3^4 = \frac{3}{2} \left( \frac{16}{2} - \frac{9}{2} \right) = \frac{3}{2} \cdot \frac{7}{2} = \frac{21}{4}
\]

For the second term:
\[
(e^2 - e) \int_3^4 1 \, dx = (e^2 - e) \cdot (4 - 3) = e^2 - e
\]

Thus, the right-hand side evaluates to:
\[
\text{RHS} = \frac{21}{4} + (e^2 - e)
\]

### Conclusion

Since the left-hand side and right-hand side are equal, the given equality is verified:
\[
\frac{21}{4} + (e^2 - e) = \frac{21}{4} + (e^2 - e)
\]

Let me know if you would like further clarification or details on any step! Here are 5 related questions for further practice:

1. What conditions must be met for Fubini's Theorem to apply?
2. How does changing the order of integration affect the limits?
3. What are some other examples where Fubini's Theorem simplifies calculations?
4. How would the problem change if the function inside the integral were discontinuous?
5. How do iterated integrals differ when applied to non-rectangular regions?

**Tip:** When solving double integrals, always check if switching the order of integration simplifies the computation!

Verify that the double integral \( \int_1^2 \left( \int_3^4 (xy + e^y) dx \right) dy = \int_3^4 \left( \int_1^2 (xy + e^y) dy \right) dx \).

Verify the equality of double integrals using Fubini's Theorem, applied to the function \(xy + e^y\). This problem explores how to switch the order of integration and evaluate iterated integrals involving both polynomials and exponential terms. Ideal for college-level calculus students.

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