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\begin{aligned} &=\int_{0}^{2}(3-x)(3-x) d x \\ &=\int_{0}^{2}\left(9-6 x+x^{2}\right) d x \\ &=9 x-3 x^{2}+\left.\frac{x^{3}}{3}\right|_{0} ^{2}=\left(18-12+\frac{8}{3}\right)-(0-0+0)=8 \frac{2}{3} \end{aligned}