1. **Problem statement:** We want to change the order of integration for the double integral $$\int_{0}^{4} \int_{0}^{y} f(x,y) \, dx \, dy$$ where the region of integration is bounded by $x=0$, $y=4$, and $y=x$. 2. **Understanding the region:** The original integral integrates $x$ from $0$ to $y$, and $y$ from $0$ to $4$. This describes the triangular region where $0 \leq x \leq y \leq 4$. 3. **Changing the order of integration:** To reverse the order, we describe $y$ in terms of $x$: - Since $x \leq y \leq 4$ and $x$ ranges from $0$ to $4$, the new limits are: - $x$ from $0$ to $4$ - $y$ from $x$ to $4$ 4. **New integral:** $$\int_{0}^{4} \int_{x}^{4} f(x,y) \, dy \, dx$$ 5. **Explanation:** Changing the order of integration means describing the same region but integrating first with respect to $y$ and then $x$. The limits reflect the same triangular region. 6. **Graph description:** The triangular region is bounded by: - The vertical line $x=0$ - The horizontal line $y=4$ - The line $y=x$ This region is the set of points $(x,y)$ such that $0 \leq x \leq y \leq 4$. **Final answer:** $$\boxed{\int_{0}^{4} \int_{0}^{y} f(x,y) \, dx \, dy = \int_{0}^{4} \int_{x}^{4} f(x,y) \, dy \, dx}$$