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Assume that $$\int_a^b g(x) \mathop{\mathrm{d}x} = \int_b^c g(x) \mathop{\mathrm{d}x}$$ for $a < b < c$.

Does this imply that

$$\int_a^b g(x) x \mathop{\mathrm{d}x} < \int_b^c g(x) x \mathop{\mathrm{d}x}$$

?

I strongly feel so but don't know how to show it.

Irregular User
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bonifaz
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    If $a,b,c$ and $g$ are positive, then this is shown true by minorating $x$ by $b$ in $\int_b^c g(x)xdx$. If they can take negative values, it is not necessarily true, take for instance $g=-1$, $a=0,b=1,c=2$. – zuggg May 21 '16 at 14:57
  • Thanks, I think that's even the answer to my question, not only a comment. – bonifaz May 21 '16 at 16:38

1 Answers1

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If $g$ is non negative, and if $\int_a^b g(x)dx>0$, then the result is true. To prove it, note that for all $x\in(a,b)$, $xg(x)<bg(x)$ whenever $g(x)>0$. Thus $$ \int_a^b xg(x)dx <b\int_a^bg(x)dx=b\int_b^cg(x)dx<\int_b^cxg(x)dx. $$ If there is no assumption on the sign of $g$, then it is not necessarily true. For instance, take $a=0,b=1,c=2$ and $g=-1$.

zuggg
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