
Yefim S. answered 07/16/21
Math Tutor with Experience
1.y2 = 4x and the line 2x + y = 4; x = y2/4; y2/2 + y = 4; y2 + 2y - 8 = 0; y = - 4 or y = 2.
Area A = ∫-42(2 - y/2 - y2/4)dy = (2y - y2/4 - y3/12)-42 = (4 - 4/4 - 8/12) - (- 8 - 4 + 64/12) = 3 - 2/3 + 12 - 16/3 = 9
2.5y2 = 16x and the curve y2 = 8x − 24. x = 5y2/16; y2 = 5y2/2 - 24; 3y2 = 48; y = ± 4
Area A = ∫-44(y2/8 + 3 - 5y2/16)dy = 2∫04(3 - 3y2/16)dy = 2(3y - y3/16)04 = 2(12 - 4) = 16