Biết $\int\limits_{-1}^{11}{f\left( x \right)}\text{d}x=18$. Tính $I=\int\limits_{0}^{2}{x\left[ 2+f\left( 3{{x}^{2}}-1 \right) \right]}\text{d}x$.

Ta có: $I=\int\limits_{0}^{2}{x\left[ 2+f\left( 3{{x}^{2}}-1 \right) \right]}\text{d}x=\int\limits_{0}^{2}{2x}\text{d}x+\int\limits_{0}^{2}{xf\left( 3{{x}^{2}}-1 \right)}\text{d}x=4+A$.
Với $A=\int\limits_{0}^{2}{xf\left( 3{{x}^{2}}-1 \right)}\text{d}x$. Đặt : $t=3{{x}^{2}}-1\Rightarrow \text{d}t=6x\text{d}x$. Lúc này: $A=\dfrac{1}{6}\int\limits_{-1}^{11}{f\left( t \right)}\text{d}t=\dfrac{1}{6}.18=3$.
Vậy: $I=4+3=7$.