Answer

问题及解答

设 $f(x)$ 在 $[a,b]$ 上可积, $\int_a^x f(x)\mathrm{d}x\geqslant 0$, $\forall x\in[a,b]$. 且 $\int_a^b f(x)\mathrm{d}x=0$. 证明: $\int_a^b xf(x)\mathrm{d}x\leqslant 0$.

Posted by haifeng on 2019-10-13 20:57:10 last update 2023-08-23 08:54:16 | Edit | Answers (1)

设 $f(x)$ 在 $[a,b]$ 上可积, $\int_a^x f(x)\mathrm{d}x\geqslant 0$, $\forall x\in[a,b]$. 且 $\int_a^b f(x)\mathrm{d}x=0$.

证明: $\int_a^b xf(x)\mathrm{d}x\leqslant 0$.

 


类似的题目, 见问题1519, 问题2854
 

1

Posted by haifeng on 2019-10-13 21:25:57

令 $\varphi(x)=\int_a^x f(t)dt$, 则结合条件, 有 $\varphi(a)=\varphi(b)=0$, 并且 $\varphi(x)\geqslant 0$, $\forall x\in[a,b]$.

$\varphi(x)\in C[a,b]$, 且在 $(a,b)$ 内可导, $\varphi'(x)=f(x)$.

\[
\int_a^b xf(x)dx=\int_a^b xd\varphi(x)=x\varphi(x)\biggr|_a^b-\int_a^b\varphi(x)dx=0-\int_a^b\varphi(x)dx\leqslant 0.
\]

 

Q.E.D.