Let I(x) be the antiderivative of x^3 (x^2 + 2)^ 1/3 with respect to x . If I( 6 ) = 24 5 , then the minimum value of I(x) for all real x is equal to

Step-by-step solution

Given I(x) = x^3 (x^2 + 2)^ 1/3 dx Let x^2 + 2 = t^3 2x dx = 3t^2 dt x dx = 3 2 t^2 dt Also, x^2 = t^3 - 2 Substituting these into the integral: I(x) = t^3 - 2 t 3 2 t^2 dt I(x) = 3 2 (t^4 - 2t) dt I(x) = 3 2 ( t^5 5 - t^2 ) + C = 3 10 t^5 - 3 2 t^2 + C We are given I( 6 ) = 24 5 . When x = 6 , t^3 = 6 + 2 = 8 t = 2 . I( 6 ) = 3 10 (2)^5 - 3 2 (2)^2 + C = 24 5 96 10 - 6 + C = 24 5 48 5 - 30 5 + C = 24 5 18 5 + C = 24 5 C = 6 5 Thus, I(x) = 3 10 (x^2 + 2)^ 5/3 - 3 2 (x^2 + 2)^ 2/3 + 6 5 To find the minimum value of