Cho \(f\left( x \right),g\left( x \right)\) là các hàm số liên tục trên R thỏa mãn \(\int\limits_0^1 {f\left( x \right)dx = 3,\int\limits_0^2 {\left[ {f\left( x \right) - 3g\left(...

* Hướng dẫn giải

Ta có: \(\int\limits_0^2 {\left[ {f\left( x \right) - 3g\left( x \right)} \right]dx = 4 \Rightarrow \int\limits_0^2 {f\left( x \right)dx - 3\int\limits_0^2 {g\left( x \right)dx = 4\,\,\,\left( 1 \right)} } } \) 

\(\int\limits_0^2 {\left[ {2f\left( x \right) + g\left( x \right)} \right]dx = 8 \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx + \int\limits_0^2 {g\left( x \right)dx = 8\,\,\,\left( 2 \right)} } } \) 

Từ (1) và (2) suy ra \(\left\{ \begin{array}{l}
\int\limits_0^2 {f\left( x \right)dx - 3\int\limits_0^2 {g\left( x \right)dx = 4} } \\
2\int\limits_0^2 {f\left( x \right)dx + \int\limits_0^2 {g\left( x \right)dx = 8} } 
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\int\limits_0^2 {f\left( x \right)dx = 4} \\
\int\limits_0^2 {g\left( x \right)dx = 0} 
\end{array} \right.\)                   

\( \Rightarrow 4 = \int\limits_0^2 {f\left( x \right)dx = \int\limits_0^1 {f\left( x \right)dx + \int\limits_1^2 {f\left( x \right)dx = 3 + \int\limits_1^2 {f\left( x \right)dx \Rightarrow \int\limits_1^2 {f\left( x \right)dx = 4 - 3 = 1} } } } } \) 

Vậy \(\int\limits_1^2 {f\left( x \right)dx = 1} \)