Section 6.1 Find the area between the curve y^{2} = 2x−2 and y = x−5. Method 1: Using dx 1.) Find points of intersection between the two curves. y^{2} = 2x − 2 and y = x − 5. (x − 5)^{2} = 2x − 2 x^{2} − 10x + 25 = 2x − 2 x^{2} − 12x + 27 = 0 (x − 3)(x − 9) = 0. Hence x = 3, 9. 2.) Determine which is larger. Between 1 and 3: \sqrt{2x − 2} > −\sqrt{2x − 2} Between 3 and 9: \sqrt{2x − 2} > x − 5 3.) Write as integral(s) Note that between 1 and 3, the height of the rectangles is \sqrt{2x − 2} − (−\sqrt{2x − 2}) and the width is dx. Note that between 3 and 9, the height of the rectangles is \sqrt{2x − 2} − (x − 5) and the width is dx. ∫_{1}^{3}[\sqrt{2x − 2} − (−\sqrt{2x − 2})]dx + ∫_{3}^{9}[\sqrt{2x − 2} − (x − 5)]dx 4.) Evaluate the integral

∫_{1}^{3}[2\sqrt{2x − 2}]dx + ∫_{3}^{9}[\sqrt{2x − 2} − (x − 5)]dx

= ∫_{1}^{3}[2\sqrt{2x − 2}]dx + ∫_{3}^{9}(\sqrt{2x − 2})dx − ∫_{3}^{9}(x − 5)dx

Let u = 2x − 2, du = 2dx, x = 1 : u = 2(1) − 2 = 0; x = 3 : u = 2(3) − 2 = 4; x = 9 : u = 2(9) − 2 = 16

= ∫_{0}^{4} u^{\frac{1}{2}} du + ∫_{4}^{16} \frac{1}{2}u^{\frac{1}{2}} du + ∫_{3}^{9}(−x + 5)dx

= \frac{2}{3}u^{\frac{3}{2}} |_{0}^{4} + \frac{1}{3}u^{\frac{3}{2}} |_{4}^{16} + (−\frac{1}{2}x^{2} + 5x)|_{3}^{9}

= \frac{2}{3}(4^{\frac{3}{2}} − 0^{\frac{3}{2}} ) + \frac{1}{3}(16^{\frac{3}{2}} − 4^{\frac{3}{2}} ) + (−\frac{1}{2}(9)^{2} + 5(9)) −(−\frac{1}{2}(3)^{2} + 5(3))

= \frac{1}{3}[2(8) + 64 − 8] − \frac{81}{2} + 45 + \frac{9}{2} − 15 = 16

= \frac{72}{3} − \frac{72}{2} + 30 = 24 − 36 + 30 = 18 Find the area between the curve y^{2} = 2x−2 and y = x−5. Method 2: Using dy 1.) Find points of intersection between the two curves. x = \frac{y^{2}+2}{2} and x = y + 5. \frac{y^{2}+2}{2} = y + 5 y^{2} + 2 = 2y + 10 y^{2} − 2y − 8 = 0. Hence, (y + 2)(y − 4) = 0, y = −2, 4. (or note from method 1, that when x = 3, y = 3 − 5 = −2 and when x = 9, y = 9 − 5 = 4 2.) Determine which function is larger. Between y = −2 and 4: when y = 0, x = \frac{y^{2}+2}{2} = \frac{0^{2}+2}{2} and x = y + 5 = 0 + 5 = 5. Hence y + 5 > \frac{y^{2}+2}{2} when y ∈ (−2, 4). 3.) Write as integral(s) Note that between -2 and 4, the height of the rectangles is y + 5 − (\frac{y^{2}+2}{2}) and the width is dy. ∫_{−2}^{4}[y + 5 − (\frac{y^{2}+2}{2})]dy 4.) Evaluate the integral: ∫_{−2}^{4}[y + 5 − (\frac{y^{2}+2}{2})]dy = ∫_{−2}^{4}[y + 5 − (\frac{y^{2}}{2} + 1)]dy = ∫_{−2}^{4}[y + 4 − \frac{y^{2}}{2} ]dy = \frac{1}{2}y^{2} + 4y − \frac{1}{6}y^{3}|_{−2}^{4} = \frac{1}{2}(4)^{2} + 4(4) − \frac{1}{6}(4)^{3} − [\frac{1}{2}(−2)^{2} + 4(−2) − \frac{1}{6}(−2)^{3}] = \frac{1}{2}(16 − 4) + 4(4 + 2) − \frac{1}{6}[64 + 8] = 6 + 24 − \frac{1}{6}[72] = 30 − 12 = 18