Câu hỏi: . Cho hàm số f(x) liên tục trên $\left[ 0;1 \right].$ Biết $\int\limits_{0}^{1}{\left[ x.{f}'\left( 1-x \right)-f\left( x \right) \right]\text{d}x}=\dfrac{1}{2},$ tính $f\left( 0 \right).$
A. $f\left( 0 \right)=-1.$
B. $f\left( 0 \right)=\dfrac{1}{2}.$
C. $f\left( 0 \right)=-\dfrac{1}{2}.$
D. $f\left( 0 \right)=1.$
A. $f\left( 0 \right)=-1.$
B. $f\left( 0 \right)=\dfrac{1}{2}.$
C. $f\left( 0 \right)=-\dfrac{1}{2}.$
D. $f\left( 0 \right)=1.$
Ta có: $I=\int\limits_{0}^{1}{\left[ x.{f}'\left( 1-x \right)-f\left( x \right) \right]dx}=\int\limits_{0}^{1}{x.{f}'\left( 1-x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}$
Đặt $t=1-x\Rightarrow dt=-dx.$ Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=0 \\
\end{aligned} \right. $, ta có $ \int\limits_{0}^{1}{x.{f}'\left( 1-x \right)dx=\int\limits_{1}^{0}{\left( 1-t \right){f}'\left( t \right)\left( -dt \right)}}$
$=\int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=1-x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$ ta có:
$\int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}=\left( 1-x \right)f\left( x \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
1 \\
\end{smallmatrix}} \right.+\int\limits_{0}^{1}{f\left( x \right)dx}=-f\left( 0 \right)+\int\limits_{0}^{1}{f\left( x \right)dx}$
Suy ra $I=-f\left( 0 \right)\Rightarrow f\left( 0 \right)=-\dfrac{1}{2}.$
Đặt $t=1-x\Rightarrow dt=-dx.$ Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=0 \\
\end{aligned} \right. $, ta có $ \int\limits_{0}^{1}{x.{f}'\left( 1-x \right)dx=\int\limits_{1}^{0}{\left( 1-t \right){f}'\left( t \right)\left( -dt \right)}}$
$=\int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=1-x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$ ta có:
$\int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}=\left( 1-x \right)f\left( x \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
1 \\
\end{smallmatrix}} \right.+\int\limits_{0}^{1}{f\left( x \right)dx}=-f\left( 0 \right)+\int\limits_{0}^{1}{f\left( x \right)dx}$
Suy ra $I=-f\left( 0 \right)\Rightarrow f\left( 0 \right)=-\dfrac{1}{2}.$
Đáp án C.