For a real number a, let I(a) = integral from -1 to 1 of (3x^2 - ax + 1) dx. Which of the following

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Question

For a real number a, let I(a) = integral from -1 to 1 of (3x^2 - ax + 1) dx. Which of the following statements is/are true?

Accepted Answer

Correct Answer: (A) and (C)Concept -- Definite Integral with Symmetry: I(a) = ∫_{-1}^{1} (3x² - a x + 1) dx = ∫_{-1}^{1} 3x² dx - a ∫_{-1}^{1} x dx + ∫_{-1}^{1} 1 dx. Since x is odd, ∫_{-1}^{1} x dx = 0. ∫_{-1}^{1} 3x² dx = [x³]_{-1}^{1} = 1 - (-1) = 2. ∫_{-1}^{1} 1 dx = 2. Thus I(a) = 2 - a·0 + 2 = 4 for all a. So I(a) is constant (positive).Why options are correct/incorrect: • A (TRUE): I(a)=4 independent of a.• B (FALSE): It does not vary.• C (TRUE): I(a)=4 > 0, so exists (in fact for all a).• D (FALSE): I(a) is always positive, never negative.Note: The integral of an odd function over a symmetric interval is zero. This property simplifies many definite integrals. Here the term -ax is odd, so it integrates to zero, leaving a constant.

For a real number a, let I(a) = integral from -1 to 1 of (3x^2 - ax + 1) dx. Which of the following statements is/are true?

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For a real number a, let I(a) = integral from -1 to 1 of (3x^2 - ax + 1) dx. Which of the following statements is/are true?

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