1. **Problem Statement:** Evaluate the double integral $$\iint_R \sin(x + y) \, dA$$ where region $R$ is bounded by $x=0$, $y=0$, and $x + y = \frac{\pi}{2}$. 2. **Understanding the Region:** The region $R$ is a right triangle in the first quadrant bounded by the coordinate axes and the line $x + y = \frac{\pi}{2}$. 3. **Setting up the Integral:** Since $x$ and $y$ are nonnegative and $x + y \leq \frac{\pi}{2}$, we can write the integral as an iterated integral: $$\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2} - x} \sin(x + y) \, dy \, dx$$ 4. **Integrate with respect to $y$ first:** $$\int_0^{\frac{\pi}{2} - x} \sin(x + y) \, dy$$ Let $u = x + y$, then $du = dy$. When $y=0$, $u=x$; when $y=\frac{\pi}{2} - x$, $u=\frac{\pi}{2}$. So the inner integral becomes: $$\int_x^{\frac{\pi}{2}} \sin u \, du = [-\cos u]_x^{\frac{\pi}{2}} = -\cos \frac{\pi}{2} + \cos x = 0 + \cos x = \cos x$$ 5. **Substitute back and integrate with respect to $x$:** $$\int_0^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_0^{\frac{\pi}{2}} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1$$ 6. **Final Answer:** $$\boxed{1}$$ This means the value of the double integral over the triangular region is 1.