Correct Answer: Option C

Explanation

\(\int^{3} _{2}(x^2 - 2x)\)dx

=\(\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{3}_{2}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{3}_{2}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\)

=\(\frac{-8}{3}\) + 4 = \(\frac{4}{3}\)

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