Lời giải của Tự Học 365

Đặt $\left\{ \begin{array}{l}u = x\\{\rm{d}}v = f'\left( x \right)\,{\rm{d}}x\end{array} \right. $ $\Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = f\left( x \right)\end{array} \right. \Rightarrow \int\limits_0^{\frac{\pi }{2}} {x.f'\left( x \right)\,{\rm{d}}x}  = \left. {x.f\left( x \right)} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\rm{d}}x} .$

Ta có $\left. {x.f\left( x \right)} \right|_0^{\frac{\pi }{2}} = \dfrac{\pi }{2}.f\left( {\dfrac{\pi }{2}} \right),$ thay $x = \dfrac{\pi }{2}$ vào giả thiết, ta được $f\left( {\dfrac{\pi }{2}} \right) + f\left( 0 \right) = 0 \Rightarrow f\left( {\dfrac{\pi }{2}} \right) = 0.$

Lại có $f\left( x \right) + f\left( {\dfrac{\pi }{2} - x} \right) = \sin x.\cos x \Leftrightarrow \int\limits_0^{\dfrac{\pi }{2}} {f\left( x \right)\,{\rm{d}}x}  + \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\dfrac{\pi }{2} - x} \right)\,{\rm{d}}x}  = \int\limits_0^{\dfrac{\pi }{2}} {\sin x.\cos x\,{\rm{d}}x} $

Đặt \(t = \dfrac{\pi }{2} - x\,\)\(\, \Rightarrow \,\,\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\text{d}}x} = \int\limits_0^{\frac{\pi }{2}} {f\left( {\dfrac{\pi }{2} - x} \right)\,{\text{d}}x} \,\, \Rightarrow \,\,\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\text{d}}x} = \dfrac{1}{4}.\)

Vậy $\int\limits_0^{\dfrac{\pi }{2}} {x.f'\left( x \right)\,{\rm{d}}x}  =  - \dfrac{1}{4}.$

Đáp án cần chọn là: d