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

To solve the problem, we start by understanding the properties of the polynomial and its roots. Given roots of \( \sqrt{3}i \) and \( 1 \), we know that the coefficients \( a, b, c \) must be real numbers, hence the third root must be the complex conjugate of \( \sqrt{3}i \), which is \( -\sqrt{3}i \).

Thus, the roots of the polynomial are \( \sqrt{3}i \), \( -\sqrt{3}i \), and \( 1 \).

Using Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots, we have:

• The sum of the roots \( \sqrt{3}i - \sqrt{3}i + 1 = 1 \) is equal to \(-a\).
• The sum of the product of the roots taken two at a time is \( 1 \times \sqrt{3}i + 1 \times (-\sqrt{3}i) + \sqrt{3}i \times (-\sqrt{3}i) = -3\), which is equal to \( b \).
• The product of the roots \( 1 \times \sqrt{3}i \times (-\sqrt{3}i) = 3 \) is equal to \(-c\).

This gives us:

• \( a = -1 \)
• \( b = -3 \)
• \( c = -3 \)

The polynomial is therefore:

\(x^3 - x^2 - 3x - 3 = 0\)

Next, we are asked to compute the integral:

\(\int_{-1}^{1} (x^3 - x^2 - 3x - 3) \, dx\)

Breaking this down:

• \(\int_{-1}^{1} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \left(\frac{1}{4} - \frac{(-1)^4}{4}\right) = 0\)
• \(\int_{-1}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \left(\frac{1}{3} - \frac{(-1)^3}{3}\right) = \frac{2}{3}\)
• \(\int_{-1}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{-1}^{1} = \left(\frac{1}{2} - \frac{(-1)^2}{2}\right) = 0\)
• \(\int_{-1}^{1} 1 \, dx = \left[x\right]_{-1}^{1} = (1 - (-1)) = 2\)

Putting it all together:

\(\int_{-1}^{1} (x^3 - x^2 - 3x - 3) \, dx = 0 - \frac{2}{3} - 3 \cdot 0 - 3 \cdot 2 = -\frac{2}{3} - 6 = -\frac{20}{3}\)

Therefore, the integral evaluates to \(-\frac{20}{3}\), making the correct answer \(-\frac{20}{3}\).