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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f\left( 5 \right)=1, \int\limits_{0}^{1}{xf\left( 5x \right)dx=1}$, khi đó $\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)dx}$ bằng
A. 15
B. 23
C. $\dfrac{123}{5}$
D. -25
A. 15
B. 23
C. $\dfrac{123}{5}$
D. -25
Đặt $t=5x\Rightarrow dt=5dx$. Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=1\Rightarrow t=5 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{1}{xf\left( 5x \right)dx}=1\Leftrightarrow \int\limits_{0}^{5}{\dfrac{t}{5}.f\left( t \right).\dfrac{dt}{5}=1}\Leftrightarrow \int\limits_{0}^{5}{t.f\left( t \right)dt}=36\Leftrightarrow \int\limits_{0}^{5}{xf\left( x \right)dx}=25$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dx={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)dx}={{x}^{2}}.f\left( x \right)\left| \begin{aligned}
& ^{5} \\
& _{0} \\
\end{aligned} \right.-2\int\limits_{0}^{5}{xf\left( x \right)dx}=25f\left( 5 \right)-2.25=-25$.
& x=0\Rightarrow t=0 \\
& x=1\Rightarrow t=5 \\
\end{aligned} \right.$
Do đó $\int\limits_{0}^{1}{xf\left( 5x \right)dx}=1\Leftrightarrow \int\limits_{0}^{5}{\dfrac{t}{5}.f\left( t \right).\dfrac{dt}{5}=1}\Leftrightarrow \int\limits_{0}^{5}{t.f\left( t \right)dt}=36\Leftrightarrow \int\limits_{0}^{5}{xf\left( x \right)dx}=25$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dx={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)dx}={{x}^{2}}.f\left( x \right)\left| \begin{aligned}
& ^{5} \\
& _{0} \\
\end{aligned} \right.-2\int\limits_{0}^{5}{xf\left( x \right)dx}=25f\left( 5 \right)-2.25=-25$.
Đáp án D.