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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$ thỏa mãn $f\left( 1 \right)=1$ và $f\left( 4x \right)-{{x}^{3}}f\left( {{x}^{4}} \right)=3{{x}^{2}}+2x+1$ với mọi $x\in \mathbb{R}.$ Khi đó $\int\limits_{1}^{4}{x.{f}'\left( x \right)\text{d}x}$ bằng
A. $I=15$.
B. $I=-1$.
C. $I=14$.
D. $I=6$.
A. $I=15$.
B. $I=-1$.
C. $I=14$.
D. $I=6$.
Ta có: $f\left( 4x \right)-{{x}^{3}}f\left( {{x}^{4}} \right)=3{{x}^{2}}+2x+1\Rightarrow 4f\left( 4x \right)-4{{x}^{3}}f\left( {{x}^{4}} \right)=12{{x}^{2}}+8x+4\left( * \right)$
$\Rightarrow 4\int\limits_{0}^{1}{f\left( 4x \right)\text{d}x-4\int\limits_{0}^{1}{{{x}^{3}}f\left( {{x}^{4}} \right)}\text{d}x=\int\limits_{0}^{1}{\left( 12{{x}^{2}}+8x+4 \right)\text{d}x=12}}$.
Đặt $4x=t\Rightarrow 4dx=dt$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\& x=1\Rightarrow t=4 \\
\end{aligned} \right.$
Nên: $4\int\limits_{0}^{4}{f\left( 4x \right)\text{d}x=\int\limits_{0}^{4}{f\left( t \right)\text{d}t}}=\int\limits_{0}^{4}{f\left( x \right)\text{d}x}$
Đặt ${{x}^{4}}=u\Rightarrow 4{{x}^{3}}\text{d}x=\text{d}u$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow u=0 \\
& x=1\Rightarrow u=1 \\
\end{aligned} \right.$
Nên: $4\int\limits_{0}^{1}{{{x}^{3}}f\left( {{x}^{4}} \right)\text{d}x=\int\limits_{0}^{1}{f\left( u \right)\text{d}u}}$.
Suy ra: $\int\limits_{0}^{4}{f\left( x \right)\text{d}x-\int\limits_{0}^{1}{f\left( x \right)\text{d}x=\int\limits_{1}^{4}{f\left( x \right)\text{d}x=12}}}$.
Đặt: $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$\Rightarrow \int\limits_{1}^{4}{x.{f}'\left( x \right)\text{d}x}=\left. xf\left( x \right) \right|_{1}^{4}-\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=4f\left( 4 \right)-f\left( 1 \right)-12=4f\left( 4 \right)-1-12=4f\left( 4 \right)-13$.
Chọn $x=1\Rightarrow 4f\left( 4 \right)-4f\left( 1 \right)=24\Leftrightarrow 4f\left( 4 \right)=28$.
Vậy $\int\limits_{1}^{4}{x{f}'\left( x \right)\text{d}x}=28-13=15$.
$\Rightarrow 4\int\limits_{0}^{1}{f\left( 4x \right)\text{d}x-4\int\limits_{0}^{1}{{{x}^{3}}f\left( {{x}^{4}} \right)}\text{d}x=\int\limits_{0}^{1}{\left( 12{{x}^{2}}+8x+4 \right)\text{d}x=12}}$.
Đặt $4x=t\Rightarrow 4dx=dt$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\& x=1\Rightarrow t=4 \\
\end{aligned} \right.$
Nên: $4\int\limits_{0}^{4}{f\left( 4x \right)\text{d}x=\int\limits_{0}^{4}{f\left( t \right)\text{d}t}}=\int\limits_{0}^{4}{f\left( x \right)\text{d}x}$
Đặt ${{x}^{4}}=u\Rightarrow 4{{x}^{3}}\text{d}x=\text{d}u$. Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow u=0 \\
& x=1\Rightarrow u=1 \\
\end{aligned} \right.$
Nên: $4\int\limits_{0}^{1}{{{x}^{3}}f\left( {{x}^{4}} \right)\text{d}x=\int\limits_{0}^{1}{f\left( u \right)\text{d}u}}$.
Suy ra: $\int\limits_{0}^{4}{f\left( x \right)\text{d}x-\int\limits_{0}^{1}{f\left( x \right)\text{d}x=\int\limits_{1}^{4}{f\left( x \right)\text{d}x=12}}}$.
Đặt: $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$\Rightarrow \int\limits_{1}^{4}{x.{f}'\left( x \right)\text{d}x}=\left. xf\left( x \right) \right|_{1}^{4}-\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=4f\left( 4 \right)-f\left( 1 \right)-12=4f\left( 4 \right)-1-12=4f\left( 4 \right)-13$.
Chọn $x=1\Rightarrow 4f\left( 4 \right)-4f\left( 1 \right)=24\Leftrightarrow 4f\left( 4 \right)=28$.
Vậy $\int\limits_{1}^{4}{x{f}'\left( x \right)\text{d}x}=28-13=15$.
Đáp án A.