# Is-a-b-f-x-dx-x-a-b-f-x-a-lt-x-lt-b-

Question Number 2776 by Filup last updated on 27/Nov/15
$$\mathrm{Is}: \\$$$$\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}\geqslant\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right) \\$$$${a}<{x}<{b} \\$$
Commented by Filup last updated on 27/Nov/15
$$\mathrm{That}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{would}\:\mathrm{assume} \\$$
Commented by Filup last updated on 27/Nov/15
$$\mathrm{This}\:\mathrm{is}\:\mathrm{interesting}.\:\mathrm{I}\:\mathrm{was}\:\mathrm{curious}\:\mathrm{and} \\$$$$\mathrm{didn}'\mathrm{t}\:\mathrm{realise}\:\mathrm{there}\:\mathrm{is}\:\mathrm{so}\:\mathrm{much}\:\mathrm{to}\:\mathrm{consider} \\$$$$\\$$$$\mathrm{I}\:\mathrm{was}\:\mathrm{thinking}\:\mathrm{more}\:\mathrm{generally}.\:\mathrm{Lets}\:\mathrm{say} \\$$$${f}\left({x}\right)={ax}^{{i}} +{bx}^{{i}−\mathrm{1}} +…+{x}^{\mathrm{1}} +{k} \\$$$$\mathrm{or}\:{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} \\$$$${or}\:{something}\:{as}\:{simple}\:{as} \\$$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\$$$$\\$$$$\\$$
Commented by Yozzi last updated on 27/Nov/15
$${What}\:{does}\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{mean}?\:{Is}\:{it}\:{implying} \\$$$${that}\:{a},{b}\in\mathbb{Z}?\:{If}\:{you}'{re}\:{trying}\:{to}\:{say} \\$$$${to}\:{add}\:{up}\:{all}\:{values}\:{of}\:{f}\left({x}\right)\:{for} \\$$$${f}\left({x}\right)\:{being}\:{continuous}\:{and}\:{positive}\:{in}\:{the} \\$$$${interval}\:{a}<{x}<{b},\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{is}\:{divergent} \\$$$${since}\:{the}\:{set}\:{X}=\left\{{x}\in\mathbb{R}\mid{a}<{x}<{b}\right\} \\$$$${has}\:{non}−{finite}\:{cardinality}.\: \\$$$${If}\:{f}\left({x}\right)\:{is}\:{continous}\:{and}\:{positive}\:{for}\:{a}<{x}<{b} \\$$$${the}\:{integral}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\:{is}\:{finite}\:{and} \\$$$${positive}.\:{The}\:{inequality}\:{would}\:{not} \\$$$${hold}. \\$$$${Alternatively},\:{f}\left({x}\right)\:{could}\:{be}\:{such}\:{that}\:{the}\:{value} \\$$$${of}\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{is}\:−\infty.\:{In}\:{this}\:{case}\:{the} \\$$$${inequality}\:{holds}\:{since}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\$$$${is}\:{finite}. \\$$
Commented by prakash jain last updated on 27/Nov/15
$$\mathrm{if}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{only}\:\mathrm{over}\:\mathrm{finite}\:\mathrm{subset}. \\$$$$\mathrm{Isn}'\mathrm{t}\:\int_{{D}} {f}\left({x}\right){dx}=\mathrm{0}\:? \\$$$$\mathrm{Assuming}\:\mid{f}\left({x}\right)\mid\:<\mathrm{M}\:\:\forall{x}\in\mathrm{D}. \\$$
Commented by 123456 last updated on 27/Nov/15
$$\mathrm{if}\:\mathrm{you}\:\mathrm{take}\:{f}\left({x}\right)=\begin{cases}{{g}\left({x}\right)\:\:\:\:{x}\in\mathrm{D}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:{x}\notin\mathrm{D}}\end{cases} \\$$$$\mathrm{where}\:\mathrm{D}\subset\left[{a},{b}\right]\:\mathrm{is}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{subset} \\$$$$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\Sigma{f}\left({x}\right)\:\mathrm{can}\:\mathrm{converge} \\$$
Commented by 123456 last updated on 27/Nov/15
$$\mathrm{yes},\:\mathrm{so}\:\mathrm{if}\:\underset{{x}\in\left[{a},{b}\right]} {\sum}{f}\left({x}\right)\:\mathrm{converge}\:\mathrm{them} \\$$$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}=\mathrm{0}\:\:\left(\mathrm{assuming}\:\mid{f}\left({x}\right)\mid\leqslant\mathrm{M},{x}\in\mathrm{D}\right) \\$$