Correct Answer: Option D

Explanation

\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)

= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)

= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]



= \(\frac{13}{3}\) + \(\frac{1}{3}\)

= \(\frac{14}{3}\)

= \(4 \frac{2}{3}\)

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