Let A _ 1 be the bounded area enclosed by the curves y=x^ 2 +2, x+y=8 and y-axis that lies in the first quadrant. Let A _ 2 be the bounded area enclosed by the curves y=x^ 2 +2, y^ 2 =x, x=2, and y-axis that lies in the first quadrant. Then A _ 1 - A _ 2 is equal to
B. 2 3 (2 2 +1)
For A_1: intersection of y = x^2 + 2 and x + y = 8 at x = 2, y = 6 A_1 = _0^2 (8-x-(x^2+2)) dx = _0^2 (6-x-x^2) dx = [6x - x^2 2 - x^3 3 ]_0^2 = 12 - 2 - 8 3 = 22 3 For A_2: bounded by y = x^2 + 2, y^2 = x (i.e., x = y^2), x = 2, and y-axis. In first quadrant, y^2 = x goes from (0,0) to (2, 2 ). The parabola y = x^2 + 2 passes through (0,2) and (2,6). A_2 = _0^ 2 (2 - y^2) dy = [2y - y^3 3 ]_0^ 2 = 2 2 - 2 2 3 = 4 2 3 A_1 - A_2 = 22 3 - 4 2 3 = 2 3 (2 2 +1)
Related: Mathematics — Area Under Curves · All PYQ Banks