\(-16\)
\(\frac{-16}{3}\)
\(\frac{16}{3}\)
\(16\)
Correct answer is C
\(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x = (\frac{8x^{1 + 1}}{2} - \frac{4x^{2+1}}{3})|_{0}^{2}\)
= \((4x^{2} - \frac{4x^{3}}{3}) |_{0}^{2}\)
= \((4(2^2) - \frac{4(2^3)}{3})\)
= \(16 - \frac{32}{3} = \frac{16}{3}\)
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