Let f(x)= d x x^ ( 2 3 ) +2 x^ ( 1 2 ) be such that f(0)=-26+24 _ e (2) . If f(1)= a + b _ e (3) , where a , b Z , then a + b is equal to :

Step-by-step solution

Let u = x^ 1/6 , so x = u^6 and dx = 6u^5 du . Then x^ 2/3 = u^4 , x^ 1/2 = u^3 . f(x) = 6u^5 du u^4 + 2u^3 = 6u^2 du u+2 . By polynomial division: u^2 u+2 = u - 2 + 4 u+2 . f(x) = 6 ( u^2 2 - 2u + 4 |u+2| ) + C = 3u^2 - 12u + 24 (u+2) + C . Substituting back: f(x) = 3x^ 1/3 - 12x^ 1/6 + 24 (x^ 1/6 +2) + C . From f(0) = -26 + 24 (2) : 24 (2) + C = -26 + 24 (2) , so C = -26 . At x = 1 : f(1) = 3 - 12 + 24 (3) - 26 = -35 + 24 (3) . Thus a = -35 , b = 24 , and a + b = -11 .