1. **Problem:** Calculate the double integral $$\iint_D (e^{xy} + x) \, dx \, dy$$ over the domain $$D = [0,1] \times [0,2].$$ 2. **Formula and rules:** For a rectangular domain $$D = [a,b] \times [c,d],$$ the double integral can be computed as an iterated integral: $$\iint_D f(x,y) \, dx \, dy = \int_c^d \int_a^b f(x,y) \, dx \, dy.$$ We can integrate term-by-term since the integral is linear. 3. **Step 1: Write the integral as iterated integral:** $$\int_0^2 \int_0^1 (e^{xy} + x) \, dx \, dy = \int_0^2 \left( \int_0^1 e^{xy} \, dx + \int_0^1 x \, dx \right) dy.$$ 4. **Step 2: Compute inner integrals:** - For $$\int_0^1 e^{xy} \, dx,$$ treat $$y$$ as constant: $$\int_0^1 e^{xy} \, dx = \left[ \frac{e^{xy}}{y} \right]_0^1 = \frac{e^y - 1}{y}$$ for $$y \neq 0$$. - For $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}.$$ 5. **Step 3: Substitute back:** $$\int_0^2 \left( \frac{e^y - 1}{y} + \frac{1}{2} \right) dy = \int_0^2 \frac{e^y - 1}{y} dy + \int_0^2 \frac{1}{2} dy.$$ 6. **Step 4: Evaluate the simpler integral:** $$\int_0^2 \frac{1}{2} dy = \frac{1}{2} \times 2 = 1.$$ 7. **Step 5: Evaluate $$\int_0^2 \frac{e^y - 1}{y} dy$$:** This integral does not have an elementary closed form but is known as the Exponential Integral related function. It can be expressed as: $$\int_0^2 \frac{e^y - 1}{y} dy = \text{Ei}(2) - \gamma - \ln 2,$$ where $$\text{Ei}$$ is the exponential integral, $$\gamma$$ is Euler-Mascheroni constant. For practical purposes, this integral can be approximated numerically. 8. **Final answer:** $$\iint_D (e^{xy} + x) \, dx \, dy = 1 + \int_0^2 \frac{e^y - 1}{y} dy.$$ This completes the solution for the first integral.