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Question Number 79837 by Pratah last updated on 28/Jan/20

Commented by mr W last updated on 28/Jan/20

$s e e Q 79814$
Commented by Pratah last updated on 28/Jan/20

$this one is different$
Commented by mr W last updated on 28/Jan/20

$i f y o u u n d e r s t a n d Q 79814, t h e n y o u$ $s h o u l d a l s o b e a b l e t o s o l v e t h i s o n e .$
Commented by Pratah last updated on 28/Jan/20

$no$
Commented by mr W last updated on 28/Jan/20

$y o u {c a n}^{'} t o r y o u {w o n}^{'} t ?$
Commented by abdomathmax last updated on 28/Jan/20

$\int_{- 1}^{1} [x - 1] d x = \int_{- 1}^{1} [x] d x - \int_{- 1}^{1} d x$ $= \int_{- 1}^{0} (- 1) d x + \int_{0}^{1}) o) d x - 2$ $= 1 - 2 = - 1$
Commented by Pratah last updated on 29/Jan/20

$no correct$
Commented by Pratah last updated on 29/Jan/20

$sir abdomathmax$
Commented by mr W last updated on 29/Jan/20

$s i r a b d o m a t h m a x :$ $e r r o r i n y o u r c a l c u l a t i o n :$ $= \int_{- 1}^{0} (- 1) d x + \int_{0}^{1} (0) d x - 2$ $= - 1 - 2 = - 3$ $i . e . y o u s h o u l d a l s o g e t t h e r e s u l t - 3 .$
Commented by mathmax by abdo last updated on 29/Jan/20

$y e s s i r t h a n k y o u .$
	Answered by key of knowledge last updated on 28/Jan/20

	$- 1 ⩽ x < 0 \Rightarrow [x - 1] = - 2 \Rightarrow \int_{- 1}^{0} (- 2) dx = - 2$ $0 ⩽ x < 1 \Rightarrow \int_{0}^{1} (- 1) dx = - 1$ $(- 1) + (- 2) = - 3$
	Commented by Pratah last updated on 28/Jan/20

	$answer : - 1$
	Commented by mr W last updated on 28/Jan/20

	$s i r P r a t a h :$ $t o u n d e r s t a n d y o u r d e f i n i t i o n, p l e a s e$ $a n s w e r :$ $[- 0.8] = - 1 o r 0 ?$
	Commented by MJS last updated on 28/Jan/20
	wait... he must answer to understand? ...or he must understand to answer? ��
	Commented by Pratah last updated on 29/Jan/20

	$[- 1 + 0.2] = - 1; ($
	Commented by mr W last updated on 29/Jan/20

	$p r a t a h s i r :$ $i f [- 0.8] = - 1$ $t h e n$ $\int_{- 1}^{1} [x - 1] d x = - 3 \neq - 1 (w h i c h y o u g a v e)$
	Commented by Pratah last updated on 29/Jan/20

	Commented by mr W last updated on 29/Jan/20

	$b u i l d y o u r o w n o p i n i o n i f y o u {d o n}^{'} t$ $h a v e o n e!$ ${t h a t}^{'} s w h y i a s k e d y o u t h e d e f i n i t i o n$ $o f [- 0.8] a t t h e b e g i n i n g!$ $y o u t o l d [- 0.8] = - 1, b u t t h i s d i a g r a m$ $s h o w s t h e g r a p h [x - 1], a n d i t s h o w s$ $a t x = 0.5 : [0.5 - 1] = [- 0.5] = 0 \neq - 1$ $a t x = - 0.5 : [- 0.5 - 1] = [- 1.5] = - 1 \neq - 2 .$ $t h a t m e a n s t h e s o l u t i o n y o u g a v e$ $u s e s a d i f f e r e n t d e f i n i t i o n f o r [x]$ $t h a n y o u!$
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