1. **State the problem:** Find the area of the shaded region bounded by the curves: - Parabola: $y = x^2$ - Line 1: $3y = 2x + 16 \implies y = \frac{2}{3}x + \frac{16}{3}$ - Line 2: $y = 8 - 2x$ The region is enclosed by points $(-2,4)$, $(1,6)$, and $(2,4)$. 2. **Identify the boundaries and intersection points:** - At $x = -2$, $y = (-2)^2 = 4$ matches the point $(-2,4)$. - At $x = 1$, check lines: - $y = \frac{2}{3}(1) + \frac{16}{3} = \frac{2}{3} + \frac{16}{3} = 6$ - $y = 8 - 2(1) = 6$ So lines intersect at $(1,6)$. - At $x = 2$, $y = 2^2 = 4$ matches $(2,4)$. 3. **Determine which curves bound the region between these points:** - Between $x = -2$ and $x = 1$, the region is bounded by parabola $y = x^2$ (bottom) and line $y = \frac{2}{3}x + \frac{16}{3}$ (top). - Between $x = 1$ and $x = 2$, bounded by parabola $y = x^2$ (bottom) and line $y = 8 - 2x$ (top). 4. **Set up the integral for the area:** $$\text{Area} = \int_{-2}^{1} \left( \frac{2}{3}x + \frac{16}{3} - x^2 \right) dx + \int_{1}^{2} \left( 8 - 2x - x^2 \right) dx$$ 5. **Calculate the first integral:** $$\int_{-2}^{1} \left( \frac{2}{3}x + \frac{16}{3} - x^2 \right) dx = \left[ \frac{1}{3}x^2 + \frac{16}{3}x - \frac{x^3}{3} \right]_{-2}^{1}$$ Evaluate at $x=1$: $$\frac{1}{3}(1)^2 + \frac{16}{3}(1) - \frac{1^3}{3} = \frac{1}{3} + \frac{16}{3} - \frac{1}{3} = \frac{16}{3}$$ Evaluate at $x=-2$: $$\frac{1}{3}(-2)^2 + \frac{16}{3}(-2) - \frac{(-2)^3}{3} = \frac{1}{3}(4) - \frac{32}{3} - \frac{-8}{3} = \frac{4}{3} - \frac{32}{3} + \frac{8}{3} = \frac{4 - 32 + 8}{3} = \frac{-20}{3}$$ Subtract: $$\frac{16}{3} - \left(-\frac{20}{3}\right) = \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$$ 6. **Calculate the second integral:** $$\int_{1}^{2} \left( 8 - 2x - x^2 \right) dx = \left[ 8x - x^2 - \frac{x^3}{3} \right]_{1}^{2}$$ Evaluate at $x=2$: $$8(2) - (2)^2 - \frac{(2)^3}{3} = 16 - 4 - \frac{8}{3} = 12 - \frac{8}{3} = \frac{36}{3} - \frac{8}{3} = \frac{28}{3}$$ Evaluate at $x=1$: $$8(1) - (1)^2 - \frac{1^3}{3} = 8 - 1 - \frac{1}{3} = 7 - \frac{1}{3} = \frac{21}{3} - \frac{1}{3} = \frac{20}{3}$$ Subtract: $$\frac{28}{3} - \frac{20}{3} = \frac{8}{3}$$ 7. **Add both areas:** $$12 + \frac{8}{3} = \frac{36}{3} + \frac{8}{3} = \frac{44}{3}$$ **Final answer:** $$\boxed{\frac{44}{3}}$$ This is the area of the shaded region.