Google
Câu hỏi: Hàm số $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ có $f\left( 0 \right)=2$ và $f\left( 4x \right)-f\left( x \right)=4{{x}^{3}}+2x, \forall x\in \mathbb{R}.$ Tích phân $I=\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{148}{63}$.
B. $\dfrac{146}{63}$.
C. $\dfrac{149}{63}$.
D. $\dfrac{145}{63}$
A. $\dfrac{148}{63}$.
B. $\dfrac{146}{63}$.
C. $\dfrac{149}{63}$.
D. $\dfrac{145}{63}$
Ta có: $\left\{ \begin{aligned}
& f\left( 4x \right)-f\left( x \right)=4{{x}^{3}}+2x, \forall x\in \mathbb{R} \\
& f\left( 0 \right)=2 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 64a-a=4 \\
& 16b-b=0 \\
& 4c-c=2 \\
& d=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{4}{63} \\
& b=0 \\
& c=\dfrac{2}{3} \\
& d=2 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=\dfrac{4}{63}{{x}^{3}}+\dfrac{2}{3}x+2$
Vậy $I=\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \dfrac{4}{63}{{x}^{3}}+\dfrac{2}{3}x+2 \right)dx}=\dfrac{148}{63}$.
& f\left( 4x \right)-f\left( x \right)=4{{x}^{3}}+2x, \forall x\in \mathbb{R} \\
& f\left( 0 \right)=2 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 64a-a=4 \\
& 16b-b=0 \\
& 4c-c=2 \\
& d=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{4}{63} \\
& b=0 \\
& c=\dfrac{2}{3} \\
& d=2 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=\dfrac{4}{63}{{x}^{3}}+\dfrac{2}{3}x+2$
Vậy $I=\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \dfrac{4}{63}{{x}^{3}}+\dfrac{2}{3}x+2 \right)dx}=\dfrac{148}{63}$.
Đáp án A.