设函数 $f(x)$ 在区间 $[0,1]$ 上连续可微, 且
$$
\int_0^1 f(x) \mathrm{d} x=3, \int_0^1 x f(x) \mathrm{d} x=3,
$$
则积分 $\int_0^1 x(x-1)\left[3-f^{\prime}(x)\right] \mathrm{d} x=$
【答案】 $\frac{5}{2}$.

【解析】
$$
\begin{aligned}
\int_0^1 x(x-1)\left[3-f^{\prime}(x)\right] \mathrm{d} x &=\int_0^1 3 x(x-1) \mathrm{d} x-\int_0^1 x(x-1) f^{\prime}(x) \mathrm{d} x \\
&=3 \times\left(\frac{1}{3}-\frac{1}{2}\right)-\int_0^1 x(x-1) \mathrm{d} f(x) \\
&=-\frac{1}{2}-\left\{\left.x(x-1) f(x)\right|_0 ^1-\int_0^1 f(x) \mathrm{d}[x(x-1)]\right\} \\
&=-\frac{1}{2}+\int_0^1 f(x)(2 x-1) \mathrm{d} x \\
&=-\frac{1}{2}+2 \int_0^1 x f(x) \mathrm{d} x-\int_0^1 f(x) \mathrm{d} x=-\frac{1}{2}+6-3=\frac{5}{2} .
\end{aligned}
$$
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