Cho hàm số f(x) liên tục trên \[\left( {0; + \infty } \right)\;\]và thỏa mãn \[2f(x) + xf\left( {\frac{1}{x}} \right) = x\;\] với mọi x>0. Tính \[\mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx\].

Ta có: \[2f\left( x \right) + xf\left( {\frac{1}{x}} \right) = x\] với\[x = \frac{1}{t}\]ta có \[2f\left( {\frac{1}{t}} \right) + \frac{1}{t}f\left( t \right) = \frac{1}{t}\]

\[ \Rightarrow f\left( {\frac{1}{t}} \right) = \frac{1}{2}\left( {\frac{1}{t} - \frac{1}{t}f\left( t \right)} \right)\]

\[ \Rightarrow f\left( {\frac{1}{x}} \right) = \frac{1}{2}\left( {\frac{1}{x} - \frac{1}{x}f\left( x \right)} \right)\]

Khi đó ta có

\[\begin{array}{*{20}{l}}{2f\left( x \right) + \frac{1}{2}x\left( {\frac{1}{x} - \frac{1}{x}f\left( x \right)} \right) = x}\\{ \Leftrightarrow 2f\left( x \right) + \frac{1}{2} - \frac{1}{2}f\left( x \right) = x}\\{ \Leftrightarrow \frac{3}{2}f\left( x \right) = x - \frac{1}{2}}\\{ \Leftrightarrow \frac{3}{2}\mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \mathop \smallint \limits_{\frac{1}{2}}^2 \left( {x - \frac{1}{2}} \right)dx}\\{ \Leftrightarrow \frac{3}{2}\mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \frac{9}{8} \Leftrightarrow \mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \frac{3}{4}}\end{array}\]