Câu 16:
Nếu ∫01f(x)dx=2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=20∫1f(x)dx=2 và ∫01g(x)dx=3\int\limits_{0}^{1}{g\left( x \right)\text{d}x}=30∫1g(x)dx=3 thì ∫01[3f(x)−2g(x)]dx\int\limits_{0}^{1}{\left[ 3f\left( x \right)-2g\left( x \right) \right]\text{d}x}0∫1[3f(x)−2g(x)]dx bằng
