If α=1 and β=1+i 2 , where i= -1 are two roots of the equation...

Since $a, b, c \in \mathbb{R}$, the complex roots of the equation $x^3+ax^2+bx+c=0$ must occur in conjugate pairs.

Given roots are $\alpha = 1$ and $\beta = 1+i\sqrt{2}$.

The third root must be $\gamma = 1-i\sqrt{2}$.

The polynomial is given by:

$x^3+ax^2+bx+c = (x-1)(x-(1+i\sqrt{2}))(x-(1-i\sqrt{2}))$

$= (x-1)((x-1)^2 - (i\sqrt{2})^2)$

$= (x-1)(x^2 - 2x + 1 + 2)$

$= (x-1)(x^2 - 2x + 3)$

$= x^3 - 3x^2 + 5x - 3$

We need to evaluate the integral:

$I = \int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) dx$

Using the property of definite integrals $\int_{-a}^{a} f(x) dx = 0$ for odd functions and $2\int_{0}^{a} f(x) dx$ for even functions:

$I = \int_{-1}^{1} x^3 dx - 3\int_{-1}^{1} x^2 dx + 5\int_{-1}^{1} x dx - 3\int_{-1}^{1} 1 dx$

$I = 0 - 3(2)\int_{0}^{1} x^2 dx + 0 - 3(2)\int_{0}^{1} 1 dx$

$I = -6 \left[ \dfrac{x^3}{3} \right]_{0}^{1} - 6 \left[ x \right]_{0}^{1}$

$I = -6 \left( \dfrac{1}{3} \right) - 6(1)$

$I = -2 - 6 = -8$

Answer: $-8$