Let A be the area of the region enclosed by the parabolas y = ax^2 - 6x and y = 2x - ax^2 , where a > 0 . If A = 16 3 , then the value of a is equal to

Step-by-step solution

To find the points of intersection, we equate the two curves: ax^2 - 6x = 2x - ax^2 2ax^2 - 8x = 0 2x(ax - 4) = 0 x = 0 or x = 4 a Since a > 0 , the upper curve in the interval [0, 4 a ] is y = 2x - ax^2 and the lower curve is y = ax^2 - 6x . The area A of the enclosed region is given by: A = ₀^ 4 a ( (2x - ax^2) - (ax^2 - 6x) ) dx A = ₀^ 4 a ( 8x - 2ax^2 ) dx Evaluating the integral: A = [ 4x^2 - 2ax^3 3 ]₀^ 4 a A = 4 ( 16 a^2 ) - 2a 3 ( 64 a^3 ) A = 64 a^2 - 128 3a^2 = 64 3a^2 We are given that A = 16 3 : 64 3a^2