Integral of Cubic Polynomial with Complex Roots

To find the value of the integral $\int_{-1}^{1} (x^3 + ax^2 + bx + c) \, dx$, we first determine the coefficients $a, b,$ and $c$ of the cubic polynomial $P(x) = x^3 + ax^2 + bx + c$.

Given that the coefficients $a, b, c \in \mathbb{R}$, any complex roots of the polynomial equation $P(x) = 0$ must occur in conjugate pairs.

We are given two roots:

1. Real root: $x_1 = \alpha = 1$
2. Complex root: $x_2 = \beta = 1 + i\sqrt{2}$

Since the coefficients are real, the third root $x_3$ must be the complex conjugate of $\beta$: $x_3 = \bar{\beta} = 1 - i\sqrt{2}$

Thus, the cubic polynomial can be factored as: $P(x) = (x - 1)(x - (1 + i\sqrt{2}))(x - (1 - i\sqrt{2}))$

Simplifying the product of the complex factors: $(x - (1 + i\sqrt{2}))(x - (1 - i\sqrt{2})) = ((x - 1) - i\sqrt{2})((x - 1) + i\sqrt{2}) = (x - 1)^2 - (i\sqrt{2})^2 = x^2 - 2x + 1 + 2 = x^2 - 2x + 3$

Now, multiplying by $(x - 1)$: $P(x) = (x - 1)(x^2 - 2x + 3)$ $P(x) = x(x^2 - 2x + 3) - 1(x^2 - 2x + 3)$ $P(x) = x^3 - 2x^2 + 3x - x^2 + 2x - 3$ $P(x) = x^3 - 3x^2 + 5x - 3$

Comparing $P(x) = x^3 + ax^2 + bx + c$, we get: $a = -3, \quad b = 5, \quad c = -3$

Now, we evaluate the definite integral: $I = \int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) \, dx$

Using the properties of definite integrals over symmetric intervals $[-a, a]$:

• The odd functions ($x^3$ and $5x$) integrate to $0$.
• The even functions ($-3x^2$ and $-3$) can be integrated by taking twice the integral from $0$ to $1$.

$I = 2 \int_{0}^{1} (-3x^2 - 3) \, dx$ $I = 2 \left[ -x^3 - 3x \right]_0^1$ $I = 2 \left( -1^3 - 3(1) - 0 \right)$ $I = 2 (-1 - 3) = 2 (-4) = -8$

Correct Option: C