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Area between curves Vertical slices. The area between the curves y = f(x) and y = g(x) over a 6 x 6 b is obtained as follows. Zb A= (f(x) − g(x)) dx a Observe that the rectangular slice has length given by f(x) − g(x) and width dx. y y = f(x) y = g(x) a b x dx Horizontal slices. The area between the curves x = f(y) and x = g(y) over c 6 y 6 d is obtained as follows. Zd A= (f(y) − g(y)) dy c Observe that the rectangular slice has length given by f(y) − g(y) and width dy. y x = g(y) x = f(y) d dy x c Gilles Cazelais. Typeset with LATEX on January 13, 2007. Example 1. Find the area bounded by the two parabolas y = x2 and y = 2x − x2 . Solution. We first find the points of intersection of the parabolas by solving x2 = 2x − x2 ⇐⇒ 2x2 − 2x = 0 ⇐⇒ 2x(x − 1) = 0. The solutions are x = 0 or 1. The points of intersections are then (0, 0) and (1, 1). y y = 2x − x2 1 y = x2 0 x 1 We find the area using vertical slices as follows. Z1 A= (2x − x2 − x2 ) dx = 0 Z1 (2x − 2x2 ) dx 0 2x3 1 = x2 − 3 0 =1− 1 2 = 3 3 Example 2. Find the area between the line y = x − 1 and the parabola y2 = 2x + 6. Solution. By solving y + 1 = (y2 − 6)/2, we can deduce that the points of intersection are (−1, −2) and (5, 4). y y2 − 6 x= 2 (5, 4) 4 x=y+1 x (−1, −2) −2 We find the area using horizontal slices as follows. 2 Z4 y −6 A= (y + 1) − dy 2 −2 4 Z4 1 y3 y2 1 2 = − 2 y + y + 4 dy = − + + 4y 2 3 2 −2 −2 1 8 = − 6 (64) + 8 + 16 − 6 + 2 − 8 = 18