1. **State the problem:** Find the area of the region bounded by the parabola $$y^{2} = 2x - 2$$ and the line $$y = x - 5$$. 2. **Rewrite the parabola equation:** $$y^{2} = 2x - 2 \implies x = \frac{y^{2} + 2}{2}$$ 3. **Rewrite the line equation in terms of $$x$$:** $$y = x - 5 \implies x = y + 5$$ 4. **Find the points of intersection by setting the $$x$$ values equal:** $$\frac{y^{2} + 2}{2} = y + 5$$ Multiply both sides by 2: $$y^{2} + 2 = 2y + 10$$ Rearranged: $$y^{2} - 2y - 8 = 0$$ 5. **Solve the quadratic equation:** $$y = \frac{2 \pm \sqrt{(-2)^{2} - 4 \cdot 1 \cdot (-8)}}{2} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$ So, $$y = \frac{2 + 6}{2} = 4$$ or $$y = \frac{2 - 6}{2} = -2$$ 6. **Determine the corresponding $$x$$ values:** For $$y=4$$: $$x = y + 5 = 4 + 5 = 9$$ For $$y=-2$$: $$x = y + 5 = -2 + 5 = 3$$ 7. **Set up the integral for the area between the curves:** The area $$A$$ is given by $$A = \int_{y=-2}^{4} \left( x_{\text{line}} - x_{\text{parabola}} \right) dy = \int_{-2}^{4} \left( (y + 5) - \frac{y^{2} + 2}{2} \right) dy$$ 8. **Simplify the integrand:** $$ (y + 5) - \frac{y^{2} + 2}{2} = y + 5 - \frac{y^{2}}{2} - 1 = -\frac{y^{2}}{2} + y + 4$$ 9. **Integrate:** $$A = \int_{-2}^{4} \left(-\frac{y^{2}}{2} + y + 4\right) dy = \left[-\frac{y^{3}}{6} + \frac{y^{2}}{2} + 4y \right]_{-2}^{4}$$ 10. **Evaluate at the bounds:** At $$y=4$$: $$-\frac{4^{3}}{6} + \frac{4^{2}}{2} + 4 \cdot 4 = -\frac{64}{6} + 8 + 16 = -\frac{32}{3} + 24 = \frac{72 - 32}{3} = \frac{40}{3}$$ At $$y=-2$$: $$-\frac{(-2)^{3}}{6} + \frac{(-2)^{2}}{2} + 4 \cdot (-2) = -\frac{-8}{6} + 2 - 8 = \frac{8}{6} + 2 - 8 = \frac{4}{3} - 6 = -\frac{14}{3}$$ 11. **Calculate the area:** $$A = \frac{40}{3} - \left(-\frac{14}{3}\right) = \frac{40}{3} + \frac{14}{3} = \frac{54}{3} = 18$$ 12. **Interpret the options:** The options are given as pairs multiplied by $$u^{2}$$. Since the area is $$18 u^{2}$$, we look for a pair whose product is 18. Check each option: - (16/3, 38/3): $$\frac{16}{3} \times \frac{38}{3} = \frac{608}{9} \approx 67.56$$ - (15/4, 35/4): $$\frac{15}{4} \times \frac{35}{4} = \frac{525}{16} = 32.8125$$ - (17/2, 37/2): $$\frac{17}{2} \times \frac{37}{2} = \frac{629}{4} = 157.25$$ - (5, 12): $$5 \times 12 = 60$$ None of these equal 18, so likely the problem expects the sum or another interpretation. Sum of pairs: - (16/3 + 38/3) = 54/3 = 18 - (15/4 + 35/4) = 50/4 = 12.5 - (17/2 + 37/2) = 54/2 = 27 - (5 + 12) = 17 The sum of (16/3, 38/3) is exactly 18, matching the area. **Final answer:** The area of the region bounded by the parabola and the line is $$18 u^{2}$$, corresponding to the pair (16/3, 38/3).