Cho f,g là hai hàm số liên tục trên [1;3] thoả mãn điều kiện $\underset{1}{\overset{3}{\int }}\left[f\left(x\right)+3g\left(x\right)\right]\text{d}x=10$ đồng thời $\underset{1}{\overset{3}{\int }}\left[2f\left(x\right)-g\left(x\right)\right]\text{d}x=6.$ Tính $\underset{1}{\overset{3}{\int }}\left[f\left(x\right)+g\left(x\right)\right]\text{d}x.$

Đáp án đúng là: A

Ta có 

$\left\{\begin{array}{l}\underset{1}{\overset{3}{\int }}\left[f\left(x\right)+3g\left(x\right)\right]\text{d}x=10\\ \underset{1}{\overset{3}{\int }}\left[2f\left(x\right)-g\left(x\right)\right]\text{d}x=6\end{array}\right.\iff \left\{\begin{array}{l}\underset{1}{\overset{3}{\int }}f\left(x\right)\text{d}x+3\underset{1}{\overset{3}{\int }}g\left(x\right)\text{d}x=10\\ 2\underset{1}{\overset{3}{\int }}f\left(x\right)\text{d}x-\underset{1}{\overset{3}{\int }}g\left(x\right)\text{d}x=6\end{array}\right.\iff \left\{\begin{array}{l}\underset{1}{\overset{3}{\int }}f\left(x\right)\text{d}x=4\\ \underset{1}{\overset{3}{\int }}g\left(x\right)\text{d}x=2\end{array}\right.$

suy ra $\underset{1}{\overset{3}{\int }}\left[f\left(x\right)+g\left(x\right)\right]\text{d}x=\underset{1}{\overset{3}{\int }}f\left(x\right)\text{d}x+\underset{1}{\overset{3}{\int }}g\left(x\right)\text{d}x=4+2=6.$