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

Ta có:  \(2f\left( x \right) + xf\left( {\dfrac{1}{x}} \right) = x\), với \(x = \dfrac{1}{t}\) ta có \(2f\left( {\dfrac{1}{t}} \right) + \dfrac{1}{t}f\left( t \right) = \dfrac{1}{t}\) \( \Rightarrow f\left( {\dfrac{1}{t}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{t} - \dfrac{1}{t}f\left( t \right)} \right)\)

\( \Rightarrow f\left( {\dfrac{1}{x}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right)\) 

Khi đó ta có

\(\begin{array}{l}2f\left( x \right) + \dfrac{1}{2}x\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right) = x\\ \Leftrightarrow 2f\left( x \right) + \dfrac{1}{2} - \dfrac{1}{2}f\left( x \right) = x\\ \Leftrightarrow \dfrac{3}{2}f\left( x \right) = x - \dfrac{1}{2}\\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{2}} \right)dx} \\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{9}{8} \Leftrightarrow \int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{3}{4}\end{array}\)