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

Is :

\int_{a}^{b} f (x) d x ⩾ \underset{x = a}{\sum^{b}} f (x)

a < x < b

Commented by Filup last updated on 27/Nov/15

That is what I would assume

Commented by Filup 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) = {a x}^{i} + {b x}^{i - 1} + \dots + x^{1} + k

or f (x) = \underset{i = 1}{\sum^{n}} a_{i} x^{i}

o r s o m e t h i n g a s s i m p l e a s

f (x) = {a x}^{2} + b x + c

Commented by Yozzi last updated on 27/Nov/15

W h a t d o e s \underset{x = a}{\sum^{b}} f (x) m e a n ? I s i t i m p l y i n g

t h a t a, b \in Z ? I f {y o u}^{'} r e t r y i n g t o s a y

t o a d d u p a l l v a l u e s o f f (x) f o r

f (x) b e i n g c o n t i n u o u s a n d p o s i t i v e i n t h e

i n t e r v a l a < x < b, \underset{x = a}{\sum^{b}} f (x) i s d i v e r g e n t

s i n c e t h e s e t X = {x \in R ∣ a < x < b}

h a s n o n - f i n i t e c a r d i n a l i t y .

I f f (x) i s c o n t i n o u s a n d p o s i t i v e f o r a < x < b

t h e i n t e g r a l \int_{a}^{b} f (x) d x i s f i n i t e a n d

p o s i t i v e . T h e i n e q u a l i t y w o u l d n o t

h o l d .

A l t e r n a t i v e l y, f (x) c o u l d b e s u c h t h a t t h e v a l u e

o f \underset{x = a}{\sum^{b}} f (x) i s - \infty . I n t h i s c a s e t h e

i n e q u a l i t y h o l d s s i n c e \int_{a}^{b} f (x) d x

i s f i n i t e .

Commented by prakash jain last updated on 27/Nov/15

if f (x) is defined only over finite subset .

{Isn}^{'} t \int_{D} f (x) d x = 0 ?

Assuming ∣ f (x) ∣ < M \forall x \in D .

Commented by 123456 last updated on 27/Nov/15

if you take f (x) = {\begin{cases} g (x) x \in D \\ 0 x \notin D \end{cases}

where D \subset [a, b] is a finite subset

i think that Σ f (x) can converge

Commented by 123456 last updated on 27/Nov/15

yes, so if \sum_{x \in [a, b]} f (x) converge them

\underset{a}{\int^{b}} f (x) d x = 0 (assuming ∣ f (x) ∣⩽ M, x \in D)

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