Biết rằng hàm số y=f(x) liên tục trên R thỏa\(f\left( 2 \right) = 5;\,\,\int\limits_0^2 {f\left( x \right)dx = \frac{4}{3}.} \) Tính\(\,I = \int\limits_0^1 {xf'\left( {2x} \right)d...

* Hướng dẫn giải

\(\begin{array}{l} \,I = \int\limits_0^1 {xf'\left( {2x} \right)dx} \\ u = f'\left( {2x} \right) \Rightarrow du = dx\\ dv = dx \Rightarrow v = \frac{1}{2}f(2x)\\ \,I = \left. {\frac{{xf(2x)}}{2}} \right|_0^1 - \frac{1}{4}\int\limits_0^1 {f\left( {2x} \right)dx} \\ = \frac{{f(2)}}{2} - \frac{1}{4}\int\limits_0^2 {f\left( x \right)dx} \\ = \frac{5}{2} - \frac{1}{4}.\frac{4}{3} = \frac{{13}}{6} \end{array}\)