Cho $f\left( x \right),g\left( x \right)$ là các hàm số liên tục trên $\mathbb{R}$ thỏa mãn $\int\limits_{0}^{1}{f\left( x \right)dx}=3,$...

Phương pháp:
Sử dụng tính chất tích phân: $\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx},\int\limits_{a}^{b}{kf\left( x \right)dx}=k\int\limits_{a}^{b}{f\left( x \right)dx}\left( k\ne 0 \right),$ $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)dx}.$
Cách giải:
Theo bài ra ta có:
$\left\{ \begin{aligned}
& \int\limits_{0}^{2}{\left[ f\left( x \right)-3g\left( x \right) \right]dx=4} \\
& \int\limits_{0}^{2}{\left[ 2f\left( x \right)+g\left( x \right) \right]dx}=8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}-3\int\limits_{0}^{2}{g\left( x \right)dx}=4 \\
& 2\int\limits_{0}^{2}{f\left( x \right)dx}+\int\limits_{0}^{2}{g\left( x \right)dx}=8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}=4 \\
& \int\limits_{0}^{2}{g\left( x \right)dx}=0 \\
\end{aligned} \right.$
Vậy $I=\int\limits_{1}^{2}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}=4-3=1.$