Question

If \(\alpha = 1\) and \(\beta = 1 + i\sqrt{2}\), where \(i = \sqrt{-1}\) are two roots of the equation
\(x^3 + ax^2 + bx + c = 0, a, b, c \in \mathbb{R}\), then \(\int_{-1}^{1} (x^3 + ax^2 + bx + c) dx\) is equal to:

• A. \(-2\)
• B. \(-4\)
• C. \(-8\)
• D. \(-10\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Since the coefficients of the polynomial are real, complex roots must occur in conjugate pairs. Thus, the third root is \(1 - i\sqrt{2}\). After finding the polynomial, we use properties of definite integration to simplify the calculation.

Step 2: Key Formula or Approach:
1. Conjugate root theorem: Roots are \(1, 1 + i\sqrt{2}, 1 - i\sqrt{2}\).
2. \(\int_{-a}^{a} f(x) dx = 0\) if \(f(x)\) is odd.

Step 3: Detailed Explanation:
The polynomial is \(f(x) = (x - 1)(x - (1 + i\sqrt{2}))(x - (1 - i\sqrt{2}))\).
\(f(x) = (x - 1)((x - 1)^2 + 2) = (x - 1)(x^2 - 2x + 1 + 2) = (x - 1)(x^2 - 2x + 3)\).
\(f(x) = x^3 - 2x^2 + 3x - x^2 + 2x - 3 = x^3 - 3x^2 + 5x - 3\).
Integrate from \(-1\) to \(1\):
\(I = \int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) dx\).
Terms \(x^3\) and \(5x\) are odd functions, so their integrals over \([-1, 1]\) are zero.
\(I = \int_{-1}^{1} (-3x^2 - 3) dx = 2 \int_{0}^{1} (-3x^2 - 3) dx \).
\(I = 2 [ -x^3 - 3x ]_0^1 = 2 [ -1 - 3 ] = -8\).

Step 4: Final Answer:
The value of the integral is \(-8\).