Area Bounded Y 32F2E3

1. **State the problem:** We need to find the area of the region bounded by the curves $u(y) = y + 7$ and $v(y) = y^2 + 6y + 7$ by integrating with respect to $y$. 2. **Identify the interval of integration:** The curves intersect at points $(2, -5)$ and $(7, 0)$, so the $y$-values range from $-5$ to $0$. 3. **Set up the integral:** The area $A$ is given by the integral of the difference between the right function and the left function along $y$: $$A = \int_{y=-5}^{0} \bigl(u(y) - v(y)\bigr) \, dy = \int_{-5}^{0} \bigl((y + 7) - (y^2 + 6y + 7)\bigr) \, dy$$ 4. **Simplify the integrand:** $$ (y + 7) - (y^2 + 6y + 7) = y + 7 - y^2 - 6y - 7 = -y^2 - 5y $$ 5. **Write the integral:** $$A = \int_{-5}^{0} (-y^2 - 5y) \, dy$$ 6. **Integrate term-by-term:** $$\int (-y^2) \, dy = -\frac{y^3}{3}$$ $$\int (-5y) \, dy = -\frac{5y^2}{2}$$ So, $$A = \left[-\frac{y^3}{3} - \frac{5y^2}{2}\right]_{-5}^{0}$$ 7. **Evaluate at the bounds:** At $y=0$: $$-\frac{0^3}{3} - \frac{5 \cdot 0^2}{2} = 0$$ At $y=-5$: $$-\frac{(-5)^3}{3} - \frac{5(-5)^2}{2} = -\frac{-125}{3} - \frac{5 \cdot 25}{2} = \frac{125}{3} - \frac{125}{2}$$ 8. **Calculate the difference:** $$A = 0 - \left(\frac{125}{3} - \frac{125}{2}\right) = -\frac{125}{3} + \frac{125}{2} = \frac{125}{2} - \frac{125}{3}$$ 9. **Find common denominator and simplify:** $$\frac{125}{2} - \frac{125}{3} = \frac{125 \cdot 3}{6} - \frac{125 \cdot 2}{6} = \frac{375 - 250}{6} = \frac{125}{6}$$ **Final answer:** $$\boxed{\frac{125}{6}}$$