Cho $\Large f,g$ là hai hàm số liên tục trên $\Large \left[ 1;3 \right

Cho $\Large f,g$ là hai hàm số liên tục trên $\Large \left[ 1;3 \right]$ thỏa mãn điều kiện $\Large \int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]}\text{dx=10}$ đồng thời $\Large \int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]}\text{dx=6}$. Tính $\Large \int\limits_{1}^{3}{f\left( 4-x \right)}\text{dx}$+$\Large 2\int\limits_{1}^{2}{g\left( 2x-1 \right)}\text{dx}$ 

• A. $\Large 9$.
• B. $\Large 6$.
• C. $\Large 7$.
• D. $\Large 8$.

Ta có: $\Large \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}$.

$\Large \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 $\Large u=\int\limits_{1}^{3}{f\left( x \right)}\text{dx; v =}\int\limits_{1}^{3}{g\left( x \right)}\text{dx}$.

Ta được hệ phương trình: $\Large \left\{ \begin{matrix}& u+3v=10 \\ & 2u-v=6 \\ \end{matrix} \right.$ $\Large \Leftrightarrow \left\{ \begin{matrix}& u=4 \\ & v=2 \\ \end{matrix} \right.$ $\Large \Rightarrow \left\{ \begin{matrix}& \int\limits_{1}^{3}{f\left( x \right)}\text{dx=4} \\ &\int\limits_{1}^{3}{g\left( x \right)}\text{dx=2} \\ \end{matrix} \right.$

+ Tính $\Large \int\limits_{1}^{3}{f\left( 4-x \right)}\text{dx}$

Đặt $\Large t=4-x\Rightarrow \text{dt}=-\text{dx};\ x=1\Rightarrow t=3;\ x=3\Rightarrow t=1$.

$\Large \int\limits_{1}^{3}{f\left( 4-x \right)\text{d}x}=\int\limits_{3}^{1}{f\left( t \right)\left( -\text{dt} \right)}=\int\limits_{1}^{3}{f\left( t \right)\text{dt}}=\int\limits_{1}^{3}{f\left( x \right)\text{dx}}=4$.

+ Tính $\Large \int\limits_{1}^{2}{g\left( 2x-1 \right)}\text{dx}$

Đặt $\Large z=2x-1\Rightarrow \text{dz}=2\text{dx};\ x=1\Rightarrow z=1;\ x=2\Rightarrow z=3$.

$\Large \int\limits_{1}^{2}{g\left( 2x-1 \right)\text{d}x}=\dfrac{1}{2}\int\limits_{1}^{3}{g\left( z \right)\text{dz}}=\dfrac{1}{2}\int\limits_{1}^{3}{g\left( x \right)\text{dx}}=1.$

Vậy $\Large \int\limits_{1}^{3}{f\left( 4-x \right)}\text{dx}$+2$\Large \int\limits_{1}^{2}{g\left( 2x-1 \right)}\text{dx = 6}$.