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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}.$ Gọi $F\left( x \right),G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 4 \right)-2G\left( 4 \right)=6$ và $F\left( -8 \right)-2G\left( -8 \right)=-2.$ Khi đó $\int\limits_{-1}^{3}{f\left( 3x-5 \right)dx}$ bằng
A. $8$.
B. $\dfrac{8}{3}$.
C. $-3$.
D. $-\dfrac{8}{3}$.
A. $8$.
B. $\dfrac{8}{3}$.
C. $-3$.
D. $-\dfrac{8}{3}$.
Đặt $3x-5=t\Rightarrow 3dx=dt$. Đổi cận: $\left\{ \begin{aligned}
& x=-1\Rightarrow t=-8 \\
& x=3\Rightarrow t=4 \\
\end{aligned} \right.$
Nên: $\int\limits_{-1}^{3}{f\left( 3x-5 \right)dx}=\dfrac{1}{3}\int\limits_{-8}^{4}{f\left( t \right)d}t=\dfrac{1}{3}\left( F\left( 4 \right)-F\left( -8 \right) \right)\text{ }\left( 1 \right)$
Vì $F\left( x \right),G\left( x \right)$ là $2$ nguyên hàm của $f\left( x \right)$ nên $F\left( x \right)=G\left( x \right)+C$.
Theo giả thiết: $\left\{ \begin{aligned}
& F\left( 4 \right)=G\left( 4 \right)+C \\
& F\left( 4 \right)=2G\left( 4 \right)+6 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2F\left( 4 \right)=2G\left( 4 \right)+2C \\
& F\left( 4 \right)=2G\left( 4 \right)+6 \\
\end{aligned} \right.\Rightarrow F\left( 4 \right)=2C-6.$
Và: $\left\{ \begin{aligned}
& F\left( -8 \right)=G\left( -8 \right)+C \\
& F\left( -8 \right)=2G\left( -8 \right)-2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2F\left( -8 \right)=2G\left( -8 \right)+2C \\
& F\left( -8 \right)=2G\left( -8 \right)-2 \\
\end{aligned} \right.\Rightarrow F\left( -8 \right)=2C+2$
Thế $F\left( 4 \right),F\left( -8 \right)$ vào $\left( 1 \right)$ ta được: $\int\limits_{-1}^{3}{f\left( 3x-5 \right)dx}=\dfrac{1}{3}\left[ \left( 2C-6 \right)-\left( 2C+2 \right) \right]=-\dfrac{8}{3}$.
& x=-1\Rightarrow t=-8 \\
& x=3\Rightarrow t=4 \\
\end{aligned} \right.$
Nên: $\int\limits_{-1}^{3}{f\left( 3x-5 \right)dx}=\dfrac{1}{3}\int\limits_{-8}^{4}{f\left( t \right)d}t=\dfrac{1}{3}\left( F\left( 4 \right)-F\left( -8 \right) \right)\text{ }\left( 1 \right)$
Vì $F\left( x \right),G\left( x \right)$ là $2$ nguyên hàm của $f\left( x \right)$ nên $F\left( x \right)=G\left( x \right)+C$.
Theo giả thiết: $\left\{ \begin{aligned}
& F\left( 4 \right)=G\left( 4 \right)+C \\
& F\left( 4 \right)=2G\left( 4 \right)+6 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2F\left( 4 \right)=2G\left( 4 \right)+2C \\
& F\left( 4 \right)=2G\left( 4 \right)+6 \\
\end{aligned} \right.\Rightarrow F\left( 4 \right)=2C-6.$
Và: $\left\{ \begin{aligned}
& F\left( -8 \right)=G\left( -8 \right)+C \\
& F\left( -8 \right)=2G\left( -8 \right)-2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2F\left( -8 \right)=2G\left( -8 \right)+2C \\
& F\left( -8 \right)=2G\left( -8 \right)-2 \\
\end{aligned} \right.\Rightarrow F\left( -8 \right)=2C+2$
Thế $F\left( 4 \right),F\left( -8 \right)$ vào $\left( 1 \right)$ ta được: $\int\limits_{-1}^{3}{f\left( 3x-5 \right)dx}=\dfrac{1}{3}\left[ \left( 2C-6 \right)-\left( 2C+2 \right) \right]=-\dfrac{8}{3}$.
Đáp án D.