[数分提高]2014-2015-2第9教学周第1次课(2015-04-28)

设 $$\bex a,b>0,\quad 0\leq f\in \calR[a,b],\quad \int_a^b xf(x)\rd x=0. \eex$$ 试证: $$\bex \int_a^b x^2f(x)\rd x\leq ab \int_a^b f(x)\rd x; \eex$$ 并给出使得下列不等式成立的 (您认为的) 最优数: $$\bex \int_a^b x^3f(x)\rd x\leq (\quad) \int_a^b f(x)\rd x. \eex$$

 

解答: 由 $$\bex 0\geq \int_a^b (x-a)(x-b)f(x)\rd x=\int_a^b x^2f(x)\rd x-ab\int_a^b f(x)\rd x \eex$$ 即知 $$\bex \int_a^b x^2f(x)\rd x\leq ab \int_a^b f(x)\rd x. \eex$$ 另外, 我们也有 $$\beex \bea &\quad \quad 0\geq \int_a^b (x-b)^3f(x)\rd x =\int_a^b x^3f(x)\rd x -3b\int_a^b x^2f(x)\rd x -b^3\int_a^b f(x)\rd x\\ &\ra \int_a^b x^3f(x)\rd x \leq 3b\cdot ab\int_a^b f(x)\rd x+b^3\int_a^b f(x)\rd x,\\ &\quad \quad 0\geq \int_a^b (x-a)^2(x-b)f(x)\rd x =\int_a^b x^3f(x)\rd x -(2a+b)\int_a^b x^2f(x)\rd x -a^2b\int_a^b f(x)\rd x\\ &\ra \int_a^b x^3f(x)\rd x \leq (2a+b)\cdot ab\int_a^b f(x)\rd x +a^2b\int_a^b f(x)\rd x. \eea \eeex$$ 于是 $$\bex \int_a^b x^3f(x)\rd x \leq \min\sed{(3a+b)b^2,(3a+b)ab}\int_a^b f(x)\rd x =(3a+b)ab\int_a^b f(x)\rd x. \eex$$