Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $f\left( 6 \right)=10$, $\int\limits_{0}^{6}{f\left( x \right)\text{d}x}=8$. Tính tích phân $I=\int\limits_{0}^{2}{x.{f}'\left( 3x \right)\text{d}x}$.
A. $I=20$.
B. $I=12$.
C. $I=52$.
D. $I=9$.
A. $I=20$.
B. $I=12$.
C. $I=52$.
D. $I=9$.
Ta có $I=\int\limits_{0}^{2}{x.{f}'\left( 3x \right)\text{d}x}=\dfrac{1}{9}\int\limits_{0}^{2}{3x.{f}'\left( 3x \right)\text{d}\left( 3x \right)}=\dfrac{1}{9}\int\limits_{0}^{6}{x.{f}'\left( x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
Suy ra $\int\limits_{0}^{6}{x.{f}'\left( x \right)\text{d}x}=\left. x.f\left( x \right) \right|_{0}^{6}-\int\limits_{0}^{6}{f\left( x \right)\text{d}x}=6f\left( 6 \right)-8=6.10-8=52$.
Vậy $I=52$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
Suy ra $\int\limits_{0}^{6}{x.{f}'\left( x \right)\text{d}x}=\left. x.f\left( x \right) \right|_{0}^{6}-\int\limits_{0}^{6}{f\left( x \right)\text{d}x}=6f\left( 6 \right)-8=6.10-8=52$.
Vậy $I=52$.
Đáp án C.