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Câu hỏi: Nếu $\int\limits_{-1}^{2}{\left[ f\left( x \right)+3g\left( x \right) \right]dx}=5$ và $\int\limits_{-1}^{2}{\left[ -f\left( x \right)+g\left( x \right) \right]dx}=-1$ thì $\int\limits_{-1}^{2}{\left[ 2f\left( x \right)+3g\left( x \right)-1 \right]dx}$ bằng:
A. 7
B. 5
C. 6
D. 4
A. 7
B. 5
C. 6
D. 4
Cách giải:
Ta có: $\left\{ \begin{aligned}
& \int\limits_{-1}^{2}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=5} \\
& \int\limits_{-1}^{2}{\left[ -f\left( x \right)+g\left( x \right) \right]dx}=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{-1}^{2}{f\left( x \right)dx}=2 \\
& \int\limits_{-1}^{2}{g\left( x \right)dx}=1 \\
\end{aligned} \right.$
Vậy $\int\limits_{-1}^{2}{\left[ 2f\left( x \right)+3g\left( x \right)-1 \right]dx}=2\int\limits_{-1}^{2}{f\left( x \right)dx}+3\int\limits_{-1}^{2}{g\left( x \right)dx}-\int\limits_{-1}^{2}{dx}=2.2+3.1-3=4.$
Ta có: $\left\{ \begin{aligned}
& \int\limits_{-1}^{2}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=5} \\
& \int\limits_{-1}^{2}{\left[ -f\left( x \right)+g\left( x \right) \right]dx}=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{-1}^{2}{f\left( x \right)dx}=2 \\
& \int\limits_{-1}^{2}{g\left( x \right)dx}=1 \\
\end{aligned} \right.$
Vậy $\int\limits_{-1}^{2}{\left[ 2f\left( x \right)+3g\left( x \right)-1 \right]dx}=2\int\limits_{-1}^{2}{f\left( x \right)dx}+3\int\limits_{-1}^{2}{g\left( x \right)dx}-\int\limits_{-1}^{2}{dx}=2.2+3.1-3=4.$
Đáp án D.