\(\frac{8}{3}\)
\(\frac{7}{3}\)
\(\frac{5}{3}\)
2
Correct answer is A
\(\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x \equiv \int_{-1}^{1} (x^{2} + 2x + 1)\mathrm {d} x\)
= \(\left. \frac{x^{3}}{3} + x^{2} + x \right |_{-1}^{1}\)
= \((\frac{1^{3}}{3} + 1^{2} + 1) - (\frac{(-1)^{3}}{3} + (-1)^{2} + (-1)) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)