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Câu hỏi: Cho $f,g$ là hai hàm số liên tục trên $\left[ 1;3 \right]$ thỏa mãn điều kiện $\int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]}\text{dx=10}$ đồng thời $\int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]}\text{dx=6}$. Tính $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right)-3{{x}^{2}} \right]}\text{dx}$.
A. $9$.
B. $-20$.
C. $6$.
D. $32$.
A. $9$.
B. $-20$.
C. $6$.
D. $32$.
Ta có: $\int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]}\text{dx=10}$ $\Leftrightarrow \int\limits_{1}^{3}{f\left( x \right)}\text{dx+3}\int\limits_{1}^{3}{g\left( x \right)}\text{dx=10}$.
$\int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]}\text{dx=6}$ $\Leftrightarrow 2\int\limits_{1}^{3}{f\left( x \right)}\text{dx-}\int\limits_{1}^{3}{g\left( x \right)}\text{dx=6}$.
Đặt $u=\int\limits_{1}^{3}{f\left( x \right)}\text{dx; v =}\int\limits_{1}^{3}{g\left( x \right)}\text{dx}$. Ta được hệ phương trình: $\left\{ \begin{aligned}
& u+3v=10 \\
& 2u-v=6 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& u=4 \\
& v=2 \\
\end{aligned} \right.$$\Rightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)}\text{dx=4} \\
& \int\limits_{1}^{3}{g\left( x \right)}\text{dx=2} \\
\end{aligned} \right.$
Vậy $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]}\text{dx=6}$ $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right)-3{{x}^{2}} \right]}\text{dx=}\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]}\text{dx-}\int\limits_{1}^{3}{3{{x}^{2}}}\text{dx=} \text{6-}\left. {{x}^{3}} \right|_{1}^{3}=6-\left( 27-1 \right)=-20$.
$\int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]}\text{dx=6}$ $\Leftrightarrow 2\int\limits_{1}^{3}{f\left( x \right)}\text{dx-}\int\limits_{1}^{3}{g\left( x \right)}\text{dx=6}$.
Đặt $u=\int\limits_{1}^{3}{f\left( x \right)}\text{dx; v =}\int\limits_{1}^{3}{g\left( x \right)}\text{dx}$. Ta được hệ phương trình: $\left\{ \begin{aligned}
& u+3v=10 \\
& 2u-v=6 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& u=4 \\
& v=2 \\
\end{aligned} \right.$$\Rightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)}\text{dx=4} \\
& \int\limits_{1}^{3}{g\left( x \right)}\text{dx=2} \\
\end{aligned} \right.$
Vậy $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]}\text{dx=6}$ $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right)-3{{x}^{2}} \right]}\text{dx=}\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]}\text{dx-}\int\limits_{1}^{3}{3{{x}^{2}}}\text{dx=} \text{6-}\left. {{x}^{3}} \right|_{1}^{3}=6-\left( 27-1 \right)=-20$.
Đáp án B.