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let $f(x),g(x)$ is continuous on $[a,b]$,and such $$\int_{a}^{x}f(t)dt\ge\int_{a}^{x}g(t)dt,x\in[a,b)$$ and $$\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt$$ show that: $$\int_{a}^{b}xf(x)dx\le\int_{a}^{b}xg(x)dx$$

my try: we only prove this $$\Longleftrightarrow \int_{a}^{b}x(f(x)-g(x))dx\le 0$$ if $[a,b]\subset (-\infty,0]$,then $x<0$ this is true.But other case I can't.Thank you

math110
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  • Integration by parts? One of the inequalities looks mixed up. – abnry Nov 29 '13 at 04:15
  • A change of variables from the negative case to the positive case also seems quite plausible, i.e. prove the negative $x$ case then change of variables from $x$ to $-x$ and use the previous result. Though, I'm sure there's a cleaner way to do this problem. – mathematics2x2life Nov 29 '13 at 04:19
  • your example is not such second condition,Thank you – math110 Nov 29 '13 at 04:19
  • I see now, in my head I forgot the minus sign in integration by parts. The problem is formulated correctly. – abnry Nov 29 '13 at 04:23

2 Answers2

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$$F (x) = \int_a^x (f-g)(x)\ dx \geq 0,\ F(a)=F(b)=0$$

$$\int_a^b x(f-g)(x)\ dx = x\int_a^x (f-g)(t)\ dt\ |_{x=a}^{x=b} - \int_a^b (1) [ \int_a^t (f-g )(s)\ ds\ ]\ dt $$ $$ = - \int_a^b F(t)\ dt \leq 0$$

HK Lee
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  • You seem to use the more compact notation of leaving out the differential symbol $dx$ (unless really necessary) in your integrals. Makes the proof look more concise.(+1) – Paramanand Singh Nov 29 '13 at 04:46
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Let $F(x) = \int_{a}^{x}f(t)\,dt, G(x) = \int_{a}^{x}g(t)\,dt$ then we have $F(x) \geq G(x)$ for all $x \in [a, b)$ and $F(b) = G(b)$. Now we have $$\int_{a}^{b}xf(x)\,dx = bF(b) - aF(a) - \int_{a}^{b}F(x)\,dx = bF(b) - \int_{a}^{b}F(x)\,dx$$ and $$\int_{a}^{b}xg(x)\,dx = bG(b) - aG(a) - \int_{a}^{b}G(x)\,dx = bG(b) - \int_{a}^{b}G(x)\,dx$$ so that

$\displaystyle \begin{aligned}\int_{a}^{b}x\{f(x) - g(x)\}\,dx &= b\{F(b) - G(b)\} - \int_{a}^{b}\{F(x) - G(x)\}\,dx\\ &= -\int_{a}^{b}\{F(x) - G(x)\}\,dx \leq 0\end{aligned}$

Paramanand Singh
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