Tích phân $\int\limits_{0}^{2}{\left( {{x}^{2}}-2x+3 \right)dx}$ bằng

Ta có $\int\limits_{0}^{2}{\left( {{x}^{2}}-2x+3 \right)\text{d}x}=\left. \dfrac{{{x}^{3}}}{3} \right|_{0}^{2}-\left. {{x}^{2}} \right|_{0}^{2}+\left. 3x \right|_{0}^{2}=\dfrac{8}{3}-4+6=\dfrac{14}{3}.$