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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 $2F\left( 11 \right)+G\left( 11 \right)=55$ và $2F\left( -1 \right)+G\left( -1 \right)=1$ Khi đó $\int\limits_{0}^{2}{x\left( 2+f\left( 3{{x}^{2}}-1 \right) \right)}\text{d}x$ bằng
A. $7$.
B. $20$.
C. $5$.
D. $22$.
A. $7$.
B. $20$.
C. $5$.
D. $22$.
Ta có $\int\limits_{0}^{2}{x\left( 2+f\left( 3{{x}^{2}}-1 \right) \right)}\text{d}x=\int\limits_{0}^{2}{2x\text{d}x}+\int\limits_{0}^{2}{xf\left( 3{{x}^{2}}-1 \right)\text{d}x}=4+\int\limits_{0}^{2}{xf\left( 3{{x}^{2}}-1 \right)\text{d}x}=4+I$.
Đặt $t=3{{x}^{2}}-1\Rightarrow dt=6xdx\Rightarrow \dfrac{1}{6}dt=xdx$.
Đổi cận: $x=0\Rightarrow t=-1; x=2\Rightarrow t=11$.
Suy ra $I=\dfrac{1}{6}\int\limits_{-1}^{11}{f\left( t \right)\text{d}t}=\dfrac{1}{6}\int\limits_{-1}^{11}{f\left( x \right)\text{d}x}=\dfrac{1}{6}\left( F\left( 11 \right)-F\left( -1 \right) \right)$.
Vì $F\left( x \right),G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ $\Rightarrow F\left( x \right)=G\left( x \right)+C$.
Suy ra $F\left( 11 \right)-F\left( -1 \right)=G\left( 11 \right)-G\left( -1 \right)$.
Ta có $\left\{ \begin{aligned}
& 2F\left( 11 \right)+G\left( 11 \right)=55 \\
& 2F\left( -1 \right)+G\left( -1 \right)=1 \\
\end{aligned} \right.\Rightarrow 2\left( F\left( 11 \right)-F\left( -1 \right) \right)+G\left( 11 \right)-G\left( -1 \right)=54$
$\Rightarrow 3\left( F\left( 11 \right)-F\left( -1 \right) \right)=54\Rightarrow F\left( 11 \right)-F\left( -1 \right)=18$.
Suy ra $I=3$. Vậy $\int\limits_{0}^{2}{x\left( 2+f\left( 3{{x}^{2}}-1 \right) \right)}\text{d}x=4+3=7$.
Đặt $t=3{{x}^{2}}-1\Rightarrow dt=6xdx\Rightarrow \dfrac{1}{6}dt=xdx$.
Đổi cận: $x=0\Rightarrow t=-1; x=2\Rightarrow t=11$.
Suy ra $I=\dfrac{1}{6}\int\limits_{-1}^{11}{f\left( t \right)\text{d}t}=\dfrac{1}{6}\int\limits_{-1}^{11}{f\left( x \right)\text{d}x}=\dfrac{1}{6}\left( F\left( 11 \right)-F\left( -1 \right) \right)$.
Vì $F\left( x \right),G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ $\Rightarrow F\left( x \right)=G\left( x \right)+C$.
Suy ra $F\left( 11 \right)-F\left( -1 \right)=G\left( 11 \right)-G\left( -1 \right)$.
Ta có $\left\{ \begin{aligned}
& 2F\left( 11 \right)+G\left( 11 \right)=55 \\
& 2F\left( -1 \right)+G\left( -1 \right)=1 \\
\end{aligned} \right.\Rightarrow 2\left( F\left( 11 \right)-F\left( -1 \right) \right)+G\left( 11 \right)-G\left( -1 \right)=54$
$\Rightarrow 3\left( F\left( 11 \right)-F\left( -1 \right) \right)=54\Rightarrow F\left( 11 \right)-F\left( -1 \right)=18$.
Suy ra $I=3$. Vậy $\int\limits_{0}^{2}{x\left( 2+f\left( 3{{x}^{2}}-1 \right) \right)}\text{d}x=4+3=7$.
Đáp án A.
