Cho hàm số f(x) liên tục trên [0;1] Biết [xf'(1-x)-f(x)]dx=1/2

Đáp án C

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 \(\left| \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 1 \Rightarrow t = 0\end{array} \right.\), 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 \(\left\{ \begin{array}{l}u = 1 - x\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = - dx\\v = f\left( x \right)\end{array} \right.\) ta có:

\(\int\limits_0^1 {\left( {1 - x} \right)f'\left( x \right)dx} = \left( {1 - x} \right)f\left( x \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle1\atop\scriptstyle}} \right. + \int\limits_0^1 {f\left( x \right)dx} = - f\left( 0 \right) + \int\limits_0^1 {f\left( x \right)dx} \)

Suy ra \(I = - f\left( 0 \right) \Rightarrow f\left( 0 \right) = - \frac{1}{2}.\)