Evaluate \(\int^{2} _{3}(x^2 - 2x)dx\)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
\(\int^{2} _{3}(x^2 - 2x)dx\\=\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{2}_{3}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{2}_{3}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\\=\frac{-8}{3}+4\\=\frac{4}{3}\)
\(\int^{2} _{3}(x^2 - 2x)dx\\=\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{2}_{3}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{2}_{3}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\\=\frac{-8}{3}+4\\=\frac{4}{3}\)