3 x+4 x^3-2 x+4 d x= f(x)+C f(3)=

Step-by-step solution

3 x+4 x^3-2 x+4 d x= f(x)+CI= 3 x+4 x^3-2 x+4 Since, x^3-2 x-4=(x-2) (x^2+2 x+2 ) 3 x+4 x^3-2 x+4 = A (x-2) + B x+C (x^2+2 x+2 ) 3 x+4=A (x^2+2 x+2 )+(B x+C)(x-2) On putting x=2, A=1 and comparing coefficients of x^2 , we get 0=A+B B=-1 On putting x=0 , we have 2 A-2 C C=-1 3 x+4 x^3-2 x+4 d x= d x x-2 - x+1 x^2+2 x+2 d x aligned & = |x-2|- 1 2 | (x^2+2 x+2 ) |+C & = f(x)+C= |x-2| (x^2+2 x+2 ) +C aligned So, f(x)= |x-2| (x^2+2 x+2 ) aligned & f(3)= |3-2| ((3)^2+2(3)+2 ) & f(3)= 1 9+6+2 = 1 17 aligned