0 < x < 1, d x x^2-x^5 = 1 3 |f(x)|+C , then f ( 1 2 )=

Step-by-step solution

d x x^2-x^5 = d x x 1-x^3 = x^2 d x x^3 1-x^3 aligned & 1-x^3=t & -3 x^2 d x=d t & -x^2 d x= -1 3 d t & - -1 3 d t (1-t) t aligned Let t =4^2 dt =24 du = -2 3 du (1+4)(1-4) = ( -2 3 ) [ 1 2 1 1+4 du + 1 2 1 1-4 du ] aligned & = -1 3 [ 1 1+4 du + 1 1-4 du ]+ c & = -1 3 [ _ e |1+4|- _ e |1-4| ]+ c & = -1 3 [ _ e |1+ t |- _ e |1- t | ]+ c & = -1 3 [ _ e |1+ 1- x ^3 |- _ e |1- 1- x ^3 | ]+ c & = 1 3 [- _ e |1+ 1- x ^3 |+ _ e |1- 1- x ^3 | ]+ c aligned = 1 3 [ _ e | 1+ 1- x ^3 1+ 1- x ^3 | ]+ c f(x)= 1- 1-x^3 1+ 1-x^3 f