1. The problem is to express the function $y = 2 + \sqrt{x+4}$ and then rearrange it to express $x$ in terms of $y$ before integrating between the bounds $y = 3$ and $y = 6$. 2. Start with the equation: $$y = 2 + \sqrt{x+4}$$ 3. Subtract 2 from both sides: $$y - 2 = \sqrt{x+4}$$ 4. Square both sides to remove the square root: $$(y - 2)^2 = x + 4$$ 5. Rearrange to solve for $x$: $$x = (y - 2)^2 - 4$$ 6. We want to integrate $x$ with respect to $y$ from $y=3$ to $y=6$: $$\int_{3}^{6} \big((y - 2)^2 - 4\big) \, dy$$ 7. Expand $(y - 2)^2$: $$(y - 2)^2 = y^2 - 4y + 4$$ 8. Substitute back: $$\int_{3}^{6} (y^2 - 4y + 4 - 4) \, dy = \int_{3}^{6} (y^2 - 4y) \, dy$$ 9. Integrate term by term: $$\int y^2 dy = \frac{y^3}{3}$$ $$\int (-4y) dy = -2y^2$$ 10. Therefore, the integral is: $$\left[ \frac{y^3}{3} - 2y^2 \right]_{3}^{6}$$ 11. Calculate at the bounds: At $y = 6$: $$\frac{6^3}{3} - 2 \times 6^2 = \frac{216}{3} - 2 \times 36 = 72 - 72 = 0$$ At $y = 3$: $$\frac{3^3}{3} - 2 \times 3^2 = \frac{27}{3} - 2 \times 9 = 9 - 18 = -9$$ 12. Find the difference to get the definite integral: $$0 - (-9) = 9$$ 13. Final answer: The value of the integral is $9$.