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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn
$xf\left( {{x}^{2}} \right)-f\left( 2x \right)=2{{x}^{3}}+2x, \forall x\in \mathbb{R}$. Tính giá trị $I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$.
A. $I=25$.
B. $I=21$.
C. $I=27$.
D. $I=23$.
$xf\left( {{x}^{2}} \right)-f\left( 2x \right)=2{{x}^{3}}+2x, \forall x\in \mathbb{R}$. Tính giá trị $I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$.
A. $I=25$.
B. $I=21$.
C. $I=27$.
D. $I=23$.
$xf\left( {{x}^{2}} \right)-f\left( 2x \right)=2{{x}^{3}}+2x\Rightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right)-f\left( 2x \right) \right]}\text{d}x=\int\limits_{1}^{2}{\left( 2{{x}^{3}}+2x \right)\text{d}x}$
$\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\left( \dfrac{{{x}^{4}}}{2}+{{x}^{2}} \right)\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\dfrac{21}{2}$. (*)
+ Tính $\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x$ :
Đặt $u={{x}^{2}}\Rightarrow \text{d}u=2x\text{d}x\Leftrightarrow x\text{d}x=\dfrac{\text{d}u}{2}$.
$x=1\Rightarrow u=1; x=2\Rightarrow u=4$.
Suy ra $\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x=\int\limits_{1}^{4}{\dfrac{f\left( u \right)}{2}\text{d}u}=\dfrac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}$.
+ Tính $\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x$ :
Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x\Leftrightarrow \text{d}x=\dfrac{\text{d}t}{2}$.
$x=1\Rightarrow t=2; x=2\Rightarrow t=4$.
Suy ra $\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\int\limits_{2}^{4}{\dfrac{f\left( t \right)}{2}\text{d}t}=\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}$.
Thay vào (*) ta được $\dfrac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}=\dfrac{21}{2}\Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}+\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x=\dfrac{21}{2}}$
$\Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\dfrac{21}{2}\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=21$.
$\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\left( \dfrac{{{x}^{4}}}{2}+{{x}^{2}} \right)\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\dfrac{21}{2}$. (*)
+ Tính $\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x$ :
Đặt $u={{x}^{2}}\Rightarrow \text{d}u=2x\text{d}x\Leftrightarrow x\text{d}x=\dfrac{\text{d}u}{2}$.
$x=1\Rightarrow u=1; x=2\Rightarrow u=4$.
Suy ra $\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x=\int\limits_{1}^{4}{\dfrac{f\left( u \right)}{2}\text{d}u}=\dfrac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}$.
+ Tính $\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x$ :
Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x\Leftrightarrow \text{d}x=\dfrac{\text{d}t}{2}$.
$x=1\Rightarrow t=2; x=2\Rightarrow t=4$.
Suy ra $\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\int\limits_{2}^{4}{\dfrac{f\left( t \right)}{2}\text{d}t}=\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}$.
Thay vào (*) ta được $\dfrac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}=\dfrac{21}{2}\Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}+\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x=\dfrac{21}{2}}$
$\Leftrightarrow \dfrac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\dfrac{21}{2}\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=21$.
Đáp án B.