1. **Problem statement:** Calculate the double integral $$I = \iint_D (x - 2y - 10) \, dx \, dy$$ where the region $$D$$ is defined by $$0 \leq x \leq 1$$ and $$0 \leq y \leq x$$. 2. **Set up the integral:** Since $$y$$ varies from $$0$$ to $$x$$ and $$x$$ varies from $$0$$ to $$1$$, the integral can be written as an iterated integral: $$I = \int_0^1 \int_0^x (x - 2y - 10) \, dy \, dx$$ 3. **Integrate with respect to $$y$$ first:** $$\int_0^x (x - 2y - 10) \, dy = \left[ xy - y^2 - 10y \right]_0^x = x \cdot x - x^2 - 10x = x^2 - x^2 - 10x = -10x$$ 4. **Substitute back and integrate with respect to $$x$$:** $$I = \int_0^1 -10x \, dx = -10 \int_0^1 x \, dx = -10 \left[ \frac{x^2}{2} \right]_0^1 = -10 \cdot \frac{1}{2} = -5$$ 5. **Final answer:** $$I = -5$$ This means the value of the double integral over the triangular region is $$-5$$.