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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f\left( 5 \right)=1$ và $\int\limits_{0}^{1}{x f\left( 5x \right)\text{d}x=1}$, khi đó $\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)dx}$ bằng
A. $-25$.
B. $23$.
C. $15$.
D. $\dfrac{123}{5}$.
A. $-25$.
B. $23$.
C. $15$.
D. $\dfrac{123}{5}$.
sTa có $\int\limits_{0}^{1}{xf\left( 5x \right)\text{d}x=1}$
Đặt $u=5x\Rightarrow \text{d}u=5\text{d}x$
Đổi cận: $x=0\Rightarrow u=0$
$x=1\Rightarrow u=5$.
Ta được $\int\limits_{0}^{1}{xf\left( 5x \right)\text{d}x=1}$ $\Rightarrow \int\limits_{0}^{5}{\dfrac{u}{5}f\left( u \right)\dfrac{\text{d}u}{5}=1\Leftrightarrow \dfrac{1}{25}\int\limits_{0}^{5}{uf\left( u \right)\text{d}u=1}}$ $\Leftrightarrow \int\limits_{0}^{5}{uf\left( u \right)\text{d}u=25}$.
Suy ra $\int\limits_{0}^{5}{xf\left( x \right)\text{d}x=25}$.
Gọi $I=\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)\text{d}x}$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2x\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$I={{x}^{2}}f\left( x \right)\left| \begin{matrix}
5 \\
0 \\
\end{matrix}-\int\limits_{0}^{5}{f\left( x \right).2x\text{d}x} \right.$$=25f\left( 5 \right)-2\int\limits_{0}^{5}{xf\left( x \right)\text{d}x} $ $ =25-2.25=-25$.
Đặt $u=5x\Rightarrow \text{d}u=5\text{d}x$
Đổi cận: $x=0\Rightarrow u=0$
$x=1\Rightarrow u=5$.
Ta được $\int\limits_{0}^{1}{xf\left( 5x \right)\text{d}x=1}$ $\Rightarrow \int\limits_{0}^{5}{\dfrac{u}{5}f\left( u \right)\dfrac{\text{d}u}{5}=1\Leftrightarrow \dfrac{1}{25}\int\limits_{0}^{5}{uf\left( u \right)\text{d}u=1}}$ $\Leftrightarrow \int\limits_{0}^{5}{uf\left( u \right)\text{d}u=25}$.
Suy ra $\int\limits_{0}^{5}{xf\left( x \right)\text{d}x=25}$.
Gọi $I=\int\limits_{0}^{5}{{{x}^{2}}{f}'\left( x \right)\text{d}x}$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2x\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$I={{x}^{2}}f\left( x \right)\left| \begin{matrix}
5 \\
0 \\
\end{matrix}-\int\limits_{0}^{5}{f\left( x \right).2x\text{d}x} \right.$$=25f\left( 5 \right)-2\int\limits_{0}^{5}{xf\left( x \right)\text{d}x} $ $ =25-2.25=-25$.
Đáp án A.