Question

For a real number \(a\), let \(I(a) = \displaystyle\int_{-1}^{1} (3x^2 - ax + 1)\, dx\). Which of the following statements is/are true?

• A. The value of \(I(a)\) is independent of the value of \(a\)
• B. The value of \(I(a)\) can vary with the value of \(a\)
• C. There exists \(a \in (-\infty, +\infty)\) such that \(I(a)\) is a positive real number
• D. There exists \(a \in (-\infty, +\infty)\) such that \(I(a)\) is a negative real number

Hint:

The term with a is an odd power of x over a symmetric interval, so it always integrates to zero; only the even-power terms survive.

Answer 1

The Correct Option is A, C

Step 1: Find the antiderivative directly.
Instead of splitting the integral into pieces first, integrate the whole polynomial $3x^2 - ax + 1$ term by term to get the antiderivative
\[ F(x) = x^3 - \frac{ax^2}{2} + x \]

Step 2: Plug in the upper limit $x=1$.
\[ F(1) = 1 - \frac{a}{2} + 1 = 2 - \frac{a}{2} \]

Step 3: Plug in the lower limit $x=-1$.
\[ F(-1) = -1 - \frac{a}{2} - 1 = -2 - \frac{a}{2} \]

Step 4: Subtract to get $I(a)$.
\[ I(a) = F(1) - F(-1) = \left(2 - \frac{a}{2}\right) - \left(-2 - \frac{a}{2}\right) \]
\[ I(a) = 2 - \frac{a}{2} + 2 + \frac{a}{2} \]
The $-\frac{a}{2}$ and $+\frac{a}{2}$ cancel out completely, leaving
\[ I(a) = 4 \]

Step 5: Read off the truth of each claim.
Since the $a$ term vanished on its own during the subtraction, the value $4$ never depends on what $a$ we pick. So claim (A) holds and claim (B) fails. Also $4$ is a fixed positive number, so we can certainly find an $a$, in fact any $a$, that makes $I(a)$ positive, which makes claim (C) true. But $I(a)$ is never negative, so claim (D) fails.

Step 6: State the answer.
\[ \boxed{I(a) = 4 \text{ for all } a} \]
The true statements are (A) and (C).