Cho \(f,\,\,g\) là hai hàm liên tục trên \(\left[ 1;3 \right]\) thỏa mãn: \(\int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=10}\) và \(\int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx}=6\). Tính \(\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]dx}\).
Trả lời bởi giáo viên
\(\begin{align} & \int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=10}\Leftrightarrow \int\limits_{1}^{3}{f(x)dx+3\int\limits_{1}^{3}{g\left( x \right)dx=10}} \\ & \int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx}=6\Leftrightarrow 2\int\limits_{1}^{3}{f\left( x \right)dx}-\int\limits_{1}^{3}{g\left( x \right)dx}=6 \\ \end{align}\)
\( \Rightarrow \left\{ \begin{align} & \int\limits_{1}^{3}{f\left( x \right)dx=}4 \\ & \int\limits_{1}^{3}{g\left( x \right)dx=2} \\ \end{align} \right.\,\,\,\,\,\,\) \(\Rightarrow \begin{align} \int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{1}^{3}{f\left( x \right)dx}+\int\limits_{1}^{3}{g\left( x \right)dx=4+2=6} \\ \end{align}\)
Hướng dẫn giải:
\(\int\limits_{a}^{b}{\left[ m.f(x)\pm n.g(x) \right]dx}=m\int\limits_{a}^{b}{f\left( x \right)dx}\pm n\int\limits_{a}^{b}{g\left( x \right)dx},\,\,\,\left( m,n\in R \right)\)