1. **Problem statement:** Evaluate the integral $$\int_0^\pi \int_x^\pi \frac{\sin y}{y} \, dy \, dx$$ and reverse the order of integration. 2. **Sketching the region:** The original integral integrates $y$ from $x$ to $\pi$ and $x$ from $0$ to $\pi$. The region in the $xy$-plane is defined by $0 \leq x \leq \pi$ and $x \leq y \leq \pi$. 3. **Reversing the order of integration:** To reverse, express $x$ in terms of $y$. Since $x \leq y \leq \pi$ and $0 \leq x \leq \pi$, for each fixed $y$, $x$ ranges from $0$ to $y$. Thus, the reversed integral is: $$\int_0^\pi \int_0^y \frac{\sin y}{y} \, dx \, dy$$ 4. **Evaluating the integral:** - The inner integral with respect to $x$ is: $$\int_0^y \frac{\sin y}{y} \, dx = \frac{\sin y}{y} \int_0^y 1 \, dx = \frac{\sin y}{y} \cdot y = \sin y$$ - So the integral reduces to: $$\int_0^\pi \sin y \, dy$$ 5. **Calculating the final integral:** $$\int_0^\pi \sin y \, dy = [-\cos y]_0^\pi = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$$ **Final answer:** $$2$$