Google
Câu hỏi: Cho hàm số $f\left( x \right)$ là hàm số có đạo hàm liên tục trên $\left[ 0;1 \right]$ và $f\left( 1 \right)=1,\int\limits_{0}^{1}{x.{f}'\left( x \right)}\text{d}x=\dfrac{2}{3}$. Tính tích phân $\int\limits_{0}^{1}{xf\left( {{x}^{2}} \right)}\text{d}x$ bằng
A. $-\dfrac{1}{6}$.
B. $-\dfrac{1}{3}$.
C. $\dfrac{1}{6}$.
D. $\dfrac{1}{3}$.
A. $-\dfrac{1}{6}$.
B. $-\dfrac{1}{3}$.
C. $\dfrac{1}{6}$.
D. $\dfrac{1}{3}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\dfrac{2}{3}=\int\limits_{0}^{1}{x.{f}'\left( x \right)}\text{d}x=x\left. f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x\Rightarrow \int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\dfrac{1}{3}$.
Ta có: $\int\limits_{0}^{1}{xf\left( {{x}^{2}} \right)}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)}\text{d}\left( {{x}^{2}} \right)=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\dfrac{1}{6}$.
& 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.$
$\dfrac{2}{3}=\int\limits_{0}^{1}{x.{f}'\left( x \right)}\text{d}x=x\left. f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x\Rightarrow \int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\dfrac{1}{3}$.
Ta có: $\int\limits_{0}^{1}{xf\left( {{x}^{2}} \right)}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)}\text{d}\left( {{x}^{2}} \right)=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\dfrac{1}{6}$.
Đáp án C.