设 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可积, 且有
$$\begin{aligned} \int_a^x f(t)\mathrm{ d} t\geq 0,\quad \int_a^b f(x)\mathrm{ d} x=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明: $\displaystyle \int_a^b xf(x)\mathrm{ d} x\leq 0$. 下述证明有瑕疵. 因为没说 $\displaystyle f$ 连续 (才能保证 $\displaystyle F'(x)=f(x)$).
$$\begin{aligned} \int_a^b xf(x)\mathrm{ d} x=&\int_a^b x\mathrm{ d} F(x)\left(F(x)\equiv \int_a^x f(t)\mathrm{ d} t\geq 0\right)\\ =&\left.xF(x)\right|_a^b -\int_a^b F(x)\mathrm{ d} x\\ =&-\int_a^b F(x)\mathrm{ d} x\leq 0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
怎么办呢? | 
发表于 2023-5-22 15:37:47
发表于 2023-5-27 09:43:57
