Cho \(\int\limits_{ - 1}^2 {f\left( x \right)dx} = 3\) và \(\int\limits_2^{ - 1} {g\left( x \right)dx} = 1\). Tính \(I = \int\limits_{ - 1}^2 {\left[ {x + 2f\left( x \right) - 3g...

* Hướng dẫn giải

\(\begin{array}{l}I = \int\limits_{ - 1}^2 {\left[ {x + 2f\left( x \right) - 3g\left( x \right)} \right]dx} \\\,\,\,\, = \int\limits_{ - 1}^2 {xdx}  + 2\int\limits_{ - 1}^2 {f\left( x \right)dx}  - 3\int\limits_{ - 1}^2 {g\left( x \right)dx} \\\,\,\,\, = \left. {\frac{{{x^2}}}{2}} \right|_{ - 1}^2 + 2\int\limits_{ - 1}^2 {f\left( x \right)dx}  + 3\int\limits_2^{ - 1} {g\left( x \right)dx} \\\,\,\,\, = \frac{{{2^2}}}{2} - \frac{{{{\left( { - 1} \right)}^2}}}{2} + 2.3 + 3.1 = \frac{{21}}{2}.\end{array}\)