1. **State the problem:** We are given the double integral $$\int_{x=-2}^{3} \int_{y=x^2+1}^{x+7} f(x,y) \, dy \, dx$$ and asked to find an equivalent integral with the order of integration reversed. 2. **Understand the region of integration:** - The outer integral is over $x$ from $-2$ to $3$. - The inner integral is over $y$ from $y = x^2 + 1$ (a parabola shifted up by 1) to $y = x + 7$ (a line). 3. **Find the range of $y$:** - At $x=-2$, $y$ ranges from $(-2)^2 + 1 = 4 + 1 = 5$ to $-2 + 7 = 5$. - At $x=3$, $y$ ranges from $3^2 + 1 = 9 + 1 = 10$ to $3 + 7 = 10$. - The minimum $y$ value on the parabola at $x=0$ is $0^2 + 1 = 1$. - The maximum $y$ value on the line at $x=3$ is $10$. So $y$ ranges from $1$ to $10$. 4. **Find $x$ in terms of $y$ for the bounds:** - From $y = x^2 + 1$, solve for $x$: $$x = \pm \sqrt{y - 1}$$ - From $y = x + 7$, solve for $x$: $$x = y - 7$$ 5. **Determine the $x$ bounds for each $y$:** - For $y$ between $1$ and $5$, the parabola $x = -\sqrt{y - 1}$ is the left boundary and the line $x = y - 7$ is the right boundary. - For $y$ between $5$ and $10$, the parabola $x = \sqrt{y - 1}$ is the right boundary and the line $x = y - 7$ is the left boundary. 6. **Write the reversed integral:** $$\int_{y=1}^{5} \int_{x=y-7}^{-\sqrt{y-1}} f(x,y) \, dx \, dy + \int_{y=5}^{10} \int_{x=y-7}^{\sqrt{y-1}} f(x,y) \, dx \, dy$$ 7. **Compare with given options:** - Option (c) matches the structure with correct bounds and order of integration. **Final answer:** $$\int_{y=1}^{5} \int_{x=y-7}^{-\sqrt{y-1}} f(x,y) \, dx \, dy + \int_{y=5}^{10} \int_{x=1-\sqrt{y-1}}^{\sqrt{y-1}} f(x,y) \, dx \, dy$$ (Note: The lower bound in the first integral is $x = y - 7$, and the upper bound is $x = -\sqrt{y-1}$, which matches the problem's structure.)