Nếu $\int\limits_{0}^{3}{f\left( x \right)dx}=6$ thì...

Ta có: $I=\int\limits_{0}^{3}{\left[ \dfrac{1}{3}f\left( x \right)+2 \right]dx}=\dfrac{1}{3}\int\limits_{0}^{3}{f\left( x \right)dx}+\int\limits_{0}^{3}{2dx}=\dfrac{1}{3} . 6+2x\left| \begin{aligned}
& 3 \\
& 0 \\
\end{aligned} \right.=8$.