The value of \(\int_{-2}^{2}\displaystyle (x^3 + 3x^2 + x)\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(16\)
On a symmetric interval \([-a, a]\), an odd function integrates to 0 and an even function gives twice the integral from 0 to a.
Here \(x^3\) and \(x\) are odd, and \(3x^2\) is even. So
\( \displaystyle \int_{-2}^{2}(x^3 + 3x^2 + x)\,dx = 0 + 2\int_{0}^{2} 3x^2\,dx + 0 \)
\( \displaystyle = 2\Big[x^3\Big]_{0}^{2} \)
\(= 2 \times 8 = 16\)
Why the other options are incorrect
Option (1) \(0\) — Incorrect. Only the odd terms give 0; the even term \(3x^2\) still contributes 16.
Option (2) \(8\) — Incorrect. This is \(\int_{0}^{2} 3x^2\,dx\) only; for an even function the integral from −2 to 2 is twice this.
Option (4) \(28\) — Incorrect. This doubles the integral of the whole expression from 0 to 2, \(2(4 + 8 + 2)\); doubling is allowed only for the even term.