Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ thoả mãn $\int\limits_{0}^{1}{x\left[ {f}'\left( x \right)-2 \right]dx=f\left( 1 \right).}$ Giá trị của $\int\limits_{0}^{1}{f\left( x \right)}dx$ bằng
A. $1.$
B. $2.$
C. $-2.$
D. $-1.$
A. $1.$
B. $2.$
C. $-2.$
D. $-1.$
$\int\limits_{0}^{1}{x\left[ {f}'\left( x \right)-2 \right]dx=\int\limits_{0}^{1}{x.{f}'\left( x \right)dx-\int\limits_{0}^{1}{2xdx}}=\int\limits_{0}^{1}{x.{f}'\left( x \right)dx-\left. {{x}^{2}} \right|_{0}^{1}}=\int\limits_{0}^{1}{x.{f}'\left( x \right)dx-1}=f\left( 1 \right).}$
$\Rightarrow \int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=f\left( 1 \right)+1.$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=\left. x.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)dx}=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx}.$
$\Rightarrow \int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx}=f\left( 1 \right)+1.$
$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=-1.$
$\Rightarrow \int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=f\left( 1 \right)+1.$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=\left. x.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)dx}=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx}.$
$\Rightarrow \int\limits_{0}^{1}{x.{f}'\left( x \right)dx}=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx}=f\left( 1 \right)+1.$
$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=-1.$
Đáp án D.