calculate the value of ∫_0 ^(1000) e^(x−[x]) dx =? where.... [x] is greatest integer functon


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Question Number 22843 by gourav~ last updated on 22/Oct/17

$c a l c u l a t e t h e v a l u e o f \int_{0}^{1000} e^{x - [x]} d x = ?$ $w h e r e . . . . [x] i s g r e a t e s t i n t e g e r f u n c t o n$
	Answered by ajfour last updated on 23/Oct/17
	$let I=∫_0 ^( 1000) e^(x−[x]) dx =Σ_0 ^(999) ∫_m ^( m+1) e^(x−[x]) dx let x−[x]={x} ⇒ dx=d{x} for each interval so I=Σ_0 ^(999) ∫_0 ^( 1) e^({x}) d{x}=1000(e−1) .$
	$l e t I = \int_{0}^{1000} e^{x - [x]} d x$ $= \underset{0}{\sum^{999}} \int_{m}^{m + 1} e^{x - [x]} d x$ $l e t x - [x] = {x}$ $\Rightarrow d x = d {x} f o r e a c h i n t e r v a l$ $s o I = \underset{0}{\sum^{999}} \int_{0}^{1} e^{{x}} d {x} = 1000 (e - 1) .$
	Commented by gourav~ last updated on 23/Oct/17

	$y e s s i r . . . b u t h o w$
	Commented by jota+ last updated on 03/Dec/17

	$m ⩽ x < m + 1 \Rightarrow x - [x] = x - m = u$ $t h e n x = m + u$ $\int_{m}^{m + 1} d x \to \int_{0}^{1} d u$
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