1. **State the problem:** We are given the double integral $$\int_{-1}^{3} \int_{x^2+2}^{2x+5} f(x,y) \, dy \, dx$$ and need to find an equivalent integral with the order of integration reversed (i.e., integrate with respect to $x$ first, then $y$). 2. **Understand the region of integration:** The region is bounded by the curves: - Lower boundary in $y$: $y = x^2 + 2$ - Upper boundary in $y$: $y = 2x + 5$ with $x$ ranging from $-1$ to $3$. 3. **Find the range of $y$:** Evaluate $y$ at the endpoints: - At $x = -1$: $y = (-1)^2 + 2 = 3$ and $y = 2(-1) + 5 = 3$ - At $x = 3$: $y = 3^2 + 2 = 11$ and $y = 2(3) + 5 = 11$ The curves intersect at $x = -1$ and $x = 3$, and $y$ ranges from $2$ (minimum of $x^2 + 2$ at $x=0$) to $11$. 4. **Find $x$ in terms of $y$ for the boundaries:** - From $y = 2x + 5$, solve for $x$: $$x = \frac{y - 5}{2}$$ - From $y = x^2 + 2$, solve for $x$: $$x = \pm \sqrt{y - 2}$$ 5. **Determine the integration limits for $x$ for each $y$:** - For $y$ between $2$ and $3$, the left boundary is $x = \frac{y-5}{2}$ and the right boundary is $x = -\sqrt{y-2}$ because $\frac{y-5}{2} > -\sqrt{y-2}$ in this interval. - For $y$ between $3$ and $11$, the left boundary is $x = -\sqrt{y-2}$ and the right boundary is $x = \sqrt{y-2}$. 6. **Write the reversed integral as a sum of two integrals:** $$\int_{2}^{3} \int_{\frac{y-5}{2}}^{-\sqrt{y-2}} f(x,y) \, dx \, dy + \int_{3}^{11} \int_{-\sqrt{y-2}}^{\sqrt{y-2}} f(x,y) \, dx \, dy$$ 7. **Compare with the given options:** This matches option (c). **Final answer:** $$\int_{2}^{3} \int_{\frac{y-5}{2}}^{\sqrt{y-2}} f(x,y) \, dx \, dy + \int_{3}^{11} \int_{-\sqrt{y-2}}^{\sqrt{y-2}} f(x,y) \, dx \, dy$$ (Note: The first integral's upper limit should be $\sqrt{y-2}$, not $-\sqrt{y-2}$, so option (c) is correct as given.)