If =1 and =1+i 2 , where i= -1 are two roots of the equation x^3+ax^2+bx+c=0 , a,b,c R , then _ -1 ¹(x^3+ax^2+bx+c)dx is equal to:

Step-by-step solution

Since a, b, c R , the complex roots of the equation x^3+ax^2+bx+c=0 must occur in conjugate pairs. Given roots are = 1 and = 1+i 2 . The third root must be = 1-i 2 . The polynomial is given by: x^3+ax^2+bx+c = (x-1)(x-(1+i 2 ))(x-(1-i 2 )) = (x-1)((x-1)^2 - (i 2 )^2) = (x-1)(x^2 - 2x + 1 + 2) = (x-1)(x^2 - 2x + 3) = x^3 - 3x^2 + 5x - 3 We need to evaluate the integral: I = _ -1 ¹ (x^3 - 3x^2 + 5x - 3) dx Using the property of definite integrals _ -a ^ a f(x) dx = 0 for odd functions and 2 ₀^ a f(x) dx for even func