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Question Number 2776 by Filup last updated on 27/Nov/15

Is: ∫_a ^( b) f(x)dx≥Σ_(x=a) ^b f(x) a<x<b

$$\mathrm{Is}: \\ $$ $$\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}\geqslant\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right) \\ $$ $${a}<{x}<{b} \\ $$

Commented byFilup last updated on 27/Nov/15

That is what I would assume

$$\mathrm{That}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{would}\:\mathrm{assume} \\ $$

Commented byFilup last updated on 27/Nov/15

This is interesting. I was curious and didn′t realise there is so much to consider I was thinking more generally. Lets say f(x)=ax^i +bx^(i−1) +...+x^1 +k or f(x)=Σ_(i=1) ^n a_i x^i or something as simple as f(x)=ax^2 +bx+c

$$\mathrm{This}\:\mathrm{is}\:\mathrm{interesting}.\:\mathrm{I}\:\mathrm{was}\:\mathrm{curious}\:\mathrm{and} \\ $$ $$\mathrm{didn}'\mathrm{t}\:\mathrm{realise}\:\mathrm{there}\:\mathrm{is}\:\mathrm{so}\:\mathrm{much}\:\mathrm{to}\:\mathrm{consider} \\ $$ $$ \\ $$ $$\mathrm{I}\:\mathrm{was}\:\mathrm{thinking}\:\mathrm{more}\:\mathrm{generally}.\:\mathrm{Lets}\:\mathrm{say} \\ $$ $${f}\left({x}\right)={ax}^{{i}} +{bx}^{{i}−\mathrm{1}} +...+{x}^{\mathrm{1}} +{k} \\ $$ $$\mathrm{or}\:{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$ $${or}\:{something}\:{as}\:{simple}\:{as} \\ $$ $${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$ $$ \\ $$ $$ \\ $$

Commented byYozzi last updated on 27/Nov/15

What does Σ_(x=a) ^b f(x) mean? Is it implying that a,b∈Z? If you′re trying to say to add up all values of f(x) for f(x) being continuous and positive in the interval a<x<b, Σ_(x=a) ^b f(x) is divergent since the set X={x∈R∣a<x<b} has non−finite cardinality. If f(x) is continous and positive for a<x<b the integral ∫_a ^b f(x)dx is finite and positive. The inequality would not hold. Alternatively, f(x) could be such that the value of Σ_(x=a) ^b f(x) is −∞. In this case the inequality holds since ∫_a ^b f(x)dx is finite.

$${What}\:{does}\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{mean}?\:{Is}\:{it}\:{implying} \\ $$ $${that}\:{a},{b}\in\mathbb{Z}?\:{If}\:{you}'{re}\:{trying}\:{to}\:{say} \\ $$ $${to}\:{add}\:{up}\:{all}\:{values}\:{of}\:{f}\left({x}\right)\:{for} \\ $$ $${f}\left({x}\right)\:{being}\:{continuous}\:{and}\:{positive}\:{in}\:{the} \\ $$ $${interval}\:{a}<{x}<{b},\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{is}\:{divergent} \\ $$ $${since}\:{the}\:{set}\:{X}=\left\{{x}\in\mathbb{R}\mid{a}<{x}<{b}\right\} \\ $$ $${has}\:{non}−{finite}\:{cardinality}.\: \\ $$ $${If}\:{f}\left({x}\right)\:{is}\:{continous}\:{and}\:{positive}\:{for}\:{a}<{x}<{b} \\ $$ $${the}\:{integral}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\:{is}\:{finite}\:{and} \\ $$ $${positive}.\:{The}\:{inequality}\:{would}\:{not} \\ $$ $${hold}. \\ $$ $${Alternatively},\:{f}\left({x}\right)\:{could}\:{be}\:{such}\:{that}\:{the}\:{value} \\ $$ $${of}\:\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right)\:{is}\:−\infty.\:{In}\:{this}\:{case}\:{the} \\ $$ $${inequality}\:{holds}\:{since}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\ $$ $${is}\:{finite}. \\ $$

Commented byprakash jain last updated on 27/Nov/15

if f(x) is defined only over finite subset. Isn′t ∫_D f(x)dx=0 ? Assuming ∣f(x)∣ <M ∀x∈D.

$$\mathrm{if}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{only}\:\mathrm{over}\:\mathrm{finite}\:\mathrm{subset}. \\ $$ $$\mathrm{Isn}'\mathrm{t}\:\int_{{D}} {f}\left({x}\right){dx}=\mathrm{0}\:? \\ $$ $$\mathrm{Assuming}\:\mid{f}\left({x}\right)\mid\:<\mathrm{M}\:\:\forall{x}\in\mathrm{D}. \\ $$

Commented by123456 last updated on 27/Nov/15

if you take f(x)= { ((g(x) x∈D)),((0 x∉D)) :} where D⊂[a,b] is a finite subset i think that Σf(x) can converge

$$\mathrm{if}\:\mathrm{you}\:\mathrm{take}\:{f}\left({x}\right)=\begin{cases}{{g}\left({x}\right)\:\:\:\:{x}\in\mathrm{D}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:{x}\notin\mathrm{D}}\end{cases} \\ $$ $$\mathrm{where}\:\mathrm{D}\subset\left[{a},{b}\right]\:\mathrm{is}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{subset} \\ $$ $$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\Sigma{f}\left({x}\right)\:\mathrm{can}\:\mathrm{converge} \\ $$

Commented by123456 last updated on 27/Nov/15

yes, so if Σ_(x∈[a,b]) f(x) converge them ∫_a ^b f(x)dx=0 (assuming ∣f(x)∣≤M,x∈D)

$$\mathrm{yes},\:\mathrm{so}\:\mathrm{if}\:\underset{{x}\in\left[{a},{b}\right]} {\sum}{f}\left({x}\right)\:\mathrm{converge}\:\mathrm{them} \\ $$ $$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}=\mathrm{0}\:\:\left(\mathrm{assuming}\:\mid{f}\left({x}\right)\mid\leqslant\mathrm{M},{x}\in\mathrm{D}\right) \\ $$

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