Nếu $\int\limits_{-2}^{1}{f\left( x \right)\text{d}x}=5$ thì...

Ta có$\int\limits_{-2}^{1}{\left[ f\left( x \right)+3 \right]\text{d}x}=\int\limits_{-2}^{1}{f\left( x \right)\text{d}x+\int\limits_{-2}^{1}{3\text{d}x}}=5+\left( 3x \right)\left| \begin{aligned}
& 1 \\
& -2 \\
\end{aligned} \right. $.$ =5+3\left( 1+2 \right)=14$