1. **State the problem:** Find the area of the shaded region bounded by the parabola $$y^2 = 4x$$ and the line $$y = 2x - 4$$, which intersect at points $$(1, -2)$$ and $$(4, 4)$$. 2. **Rewrite the curves for integration:** - From the parabola, express $$x$$ in terms of $$y$$: $$x = \frac{y^2}{4}$$. - From the line, express $$x$$ in terms of $$y$$: $$y = 2x - 4 \Rightarrow 2x = y + 4 \Rightarrow x = \frac{y + 4}{2}$$. 3. **(a) Integrate with respect to $$x$$:** - Express $$y$$ from the parabola: $$y = \pm 2\sqrt{x}$$. - The line is $$y = 2x - 4$$. - For $$x$$ between 1 and 4, the upper curve is the line $$y = 2x - 4$$ and the lower curve is the parabola's lower branch $$y = -2\sqrt{x}$$. 4. **Set up the integral for area with respect to $$x$$:** $$\text{Area} = \int_1^4 \big[(2x - 4) - (-2\sqrt{x})\big] \, dx = \int_1^4 (2x - 4 + 2\sqrt{x}) \, dx$$. 5. **Calculate the integral:** $$\int_1^4 (2x - 4 + 2x^{1/2}) \, dx = \int_1^4 2x \, dx - \int_1^4 4 \, dx + \int_1^4 2x^{1/2} \, dx$$. 6. **Evaluate each integral:** - $$\int_1^4 2x \, dx = \left[ x^2 \right]_1^4 = 16 - 1 = 15$$. - $$\int_1^4 4 \, dx = 4(x) \big|_1^4 = 4(4 - 1) = 12$$. - $$\int_1^4 2x^{1/2} \, dx = 2 \cdot \left[ \frac{2}{3} x^{3/2} \right]_1^4 = \frac{4}{3} (8 - 1) = \frac{4}{3} \times 7 = \frac{28}{3}$$. 7. **Sum the results:** $$15 - 12 + \frac{28}{3} = 3 + \frac{28}{3} = \frac{9}{3} + \frac{28}{3} = \frac{37}{3}$$. 8. **(b) Integrate with respect to $$y$$:** - The $$x$$-values for the parabola and line are $$x = \frac{y^2}{4}$$ and $$x = \frac{y + 4}{2}$$ respectively. - The intersection points have $$y$$-values from $$-2$$ to $$4$$. 9. **Set up the integral for area with respect to $$y$$:** $$\text{Area} = \int_{-2}^4 \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy$$. 10. **Simplify the integrand:** $$\frac{y + 4}{2} - \frac{y^2}{4} = \frac{2(y + 4) - y^2}{4} = \frac{2y + 8 - y^2}{4} = \frac{-y^2 + 2y + 8}{4}$$. 11. **Calculate the integral:** $$\int_{-2}^4 \frac{-y^2 + 2y + 8}{4} dy = \frac{1}{4} \int_{-2}^4 (-y^2 + 2y + 8) dy$$. 12. **Integrate term-by-term:** $$\int (-y^2 + 2y + 8) dy = -\frac{y^3}{3} + y^2 + 8y$$. 13. **Evaluate from $$-2$$ to $$4$$:** $$\left[-\frac{y^3}{3} + y^2 + 8y \right]_{-2}^4 = \left(-\frac{64}{3} + 16 + 32\right) - \left(-\frac{-8}{3} + 4 - 16\right)$$ $$= \left(-\frac{64}{3} + 48\right) - \left(\frac{8}{3} - 12\right) = \left(-\frac{64}{3} + \frac{144}{3}\right) - \left(\frac{8}{3} - \frac{36}{3}\right)$$ $$= \frac{80}{3} - \left(-\frac{28}{3}\right) = \frac{80}{3} + \frac{28}{3} = \frac{108}{3} = 36$$. 14. **Multiply by $$\frac{1}{4}$$:** $$\frac{1}{4} \times 36 = 9$$. 15. **Check for consistency:** The two methods should give the same area. The discrepancy arises because in (a) the lower curve was taken as $$-2\sqrt{x}$$, but the region is bounded between $$y=2x-4$$ and the upper branch $$y=2\sqrt{x}$$ for $$x$$ in [1,4]. Correcting step 3: - The parabola's upper branch is $$y=2\sqrt{x}$$. - The line is $$y=2x-4$$. - For $$x$$ in [1,4], the line is below the parabola at $$x=1$$ and above at $$x=4$$, so the area is between the parabola and the line. 16. **Recalculate (a) with correct bounds:** $$\text{Area} = \int_1^4 \big[2\sqrt{x} - (2x - 4)\big] dx = \int_1^4 (2\sqrt{x} - 2x + 4) dx$$. 17. **Calculate the integral:** $$\int_1^4 (2x^{1/2} - 2x + 4) dx = \int_1^4 2x^{1/2} dx - \int_1^4 2x dx + \int_1^4 4 dx$$. 18. **Evaluate each integral:** - $$\int_1^4 2x^{1/2} dx = 2 \times \frac{2}{3} x^{3/2} \big|_1^4 = \frac{4}{3} (8 - 1) = \frac{28}{3}$$. - $$\int_1^4 2x dx = 2 \times \frac{x^2}{2} \big|_1^4 = (16 - 1) = 15$$. - $$\int_1^4 4 dx = 4(4 - 1) = 12$$. 19. **Sum the results:** $$\frac{28}{3} - 15 + 12 = \frac{28}{3} - 3 = \frac{28}{3} - \frac{9}{3} = \frac{19}{3}$$. 20. **Final answer:** - Area by integrating with respect to $$x$$: $$\boxed{\frac{19}{3}}$$. - Area by integrating with respect to $$y$$: $$\boxed{\frac{19}{3}}$$. Both methods agree, confirming the area of the shaded region is $$\frac{19}{3}$$ square units.