Question

If \(\sqrt{3}i, 1\) are the roots of \(x^{3} + ax^{2} + bx + c = 0\) where \(a, b, c \in \mathbb{R}\). Then \(\int_{-1}^{1} (x^{3} + ax^{2} + bx + c)dx\) is

• A. \(\frac{20}{3}\)
• B. \(\frac{10}{3}\)
• C. \(-\frac{20}{3}\)
• D. \(-\frac{10}{3}\)

Hint:

For integrals over symmetric intervals \([-a, a]\), always separate the function into odd and even parts. Integrating only the even part and doubling it while setting the odd part to zero significantly reduces calculation errors and saves time.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two roots of a cubic polynomial with real coefficients. Using the property that complex roots occur in conjugate pairs, we can find the third root, reconstruct the polynomial, and then compute its definite integral over a symmetric interval.
Step 2: Key Formula or Approach: 1. If \(x+iy\) is a root of a polynomial with real coefficients, then \(x-iy\) is also a root.
2. \(\int_{-a}^{a} f(x) dx = 0\) if \(f(x)\) is an odd function.
3. \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\) if \(f(x)\) is an even function.
Step 3: Detailed Explanation:
The given roots are \(1\) and \(\sqrt{3}i\). Since the coefficients \(a, b, c\) are real, the complex conjugate of \(\sqrt{3}i\), which is \(-\sqrt{3}i\), must also be a root.
The polynomial \(P(x)\) can be written as:
\[ P(x) = (x - 1)(x - \sqrt{3}i)(x + \sqrt{3}i) \]
\[ P(x) = (x - 1)(x^{2} + 3) = x^{3} - x^{2} + 3x - 3 \]
Comparing with \(x^{3} + ax^{2} + bx + c = 0\), we get \(a = -1, b = 3, c = -3\).
The integral to evaluate is:
\[ I = \int_{-1}^{1} (x^{3} - x^{2} + 3x - 3) dx \]
Using the properties of integrals for odd and even functions:
\[ I = \int_{-1}^{1} x^{3} dx - \int_{-1}^{1} x^{2} dx + \int_{-1}^{1} 3x dx - \int_{-1}^{1} 3 dx \]
Since \(x^{3}\) and \(3x\) are odd functions, their integrals from \(-1\) to \(1\) are zero.
\[ I = 0 - 2 \int_{0}^{1} x^{2} dx + 0 - 2 \int_{0}^{1} 3 dx \]
\[ I = -2 \left[ \frac{x^{3}}{3} \right]_{0}^{1} - 2 [3x]_{0}^{1} \]
\[ I = -2 \left( \frac{1}{3} \right) - 2(3) = -\frac{2}{3} - 6 = -\frac{20}{3} \]
Step 4: Final Answer:
The value of the integral is \(-\frac{20}{3}\), which corresponds to option (C).