Cho \(\int\limits_{1}^{2}{\left[ 4f\left( x \right)-2x \right]dx}=1\). Khi đó \(\int\limits_{1}{2}{f\left( x \right)}dx\)

* Hướng dẫn giải

\(\begin{array}{l} \int\limits_1^2 {\left[ {4f\left( x \right) - 2x} \right]dx} = 1 \Leftrightarrow 4\int\limits_1^2 {f\left( x \right)dx - 2\int\limits_1^2 {xdx} } = 1 \Leftrightarrow 4\int\limits_1^2 {f\left( x \right)dx - 2.} \left. {\frac{{{x^2}}}{2}} \right|_1^2 = 1\\ \Leftrightarrow 4\int\limits_1^2 {f\left( x \right)dx = 4 \Leftrightarrow } \int\limits_1^2 {f\left( x \right)dx = 1} \end{array}\)