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Question Number 52161 by naka3546 last updated on 03/Jan/19

Commented by prakash jain last updated on 04/Jan/19
$f(x)= { ((7x+1),(0≤x≤1/2)),((7−5x),(1/2≤x≤1)) :} ∫_0 ^1 f(x)dx=∫_0 ^(1/2) f(x)dx+∫_(1 /2) ^1 f(x)dx =[((7x^2 )/2)+x]_0 ^(1/2) +[7x−((5x^2 )/2)]_(1/2) ^1 =((7/8)+(1/2))+(7−(5/2))−((7/2)−(5/8)) =((7+4+56−20−28+5)/8)=((24)/3)=3 f(x) above satifies all conditons. f(0)=1 f(1)=2 ∫_0 ^1 f(x)dx=3 xf(x)>0 for 0<x<1 ∫_0 ^1 xf(x)dx is unlikely to be 0. You can re−check. I will my workinhs again.$
$f (x) = {\begin{cases} 7 x + 1 & 0 ⩽ x ⩽ 1 / 2 \\ 7 - 5 x & 1 / 2 ⩽ x ⩽ 1 \end{cases}$ $\int_{0}^{1} f (x) d x = \int_{0}^{1 / 2} f (x) d x + \int_{1 / 2}^{1} f (x) d x$ $= {[\frac{7 x^{2}}{2} + x]}_{0}^{1 / 2} + {[7 x - \frac{5 x^{2}}{2}]}_{1 / 2}^{1}$ $= (\frac{7}{8} + \frac{1}{2}) + (7 - \frac{5}{2}) - (\frac{7}{2} - \frac{5}{8})$ $= \frac{7 + 4 + 56 - 20 - 28 + 5}{8} = \frac{24}{3} = 3$ $f (x) a b o v e s a t i f i e s a l l c o n d i t o n s .$ $f (0) = 1$ $f (1) = 2$ $\int_{0}^{1} f (x) d x = 3$ $x f (x) > 0 f o r 0 < x < 1$ $\int_{0}^{1} x f (x) d x is unlikely to be 0 .$ $You can re - check . I will my workinhs$ $again .$
Commented by mr W last updated on 04/Jan/19

$\int_{0}^{1} f (x) d x r e p r e s e n t s t h e a r e a . t h e$ $a r e a i s 3 . b u t i n f i n i t e l y m a n y s h a p e s$ $c a n h a v e t h e s a m e a r e a .$ $\int_{0}^{1} x f (x) d x r e p r e s e n t s t h e s t a t i c a l m o m e n t$ $o f t h e a r e a a b o u t y - a x i s . i t d e p e n d s$ $n o t o n l y o n t h e s i z e o f t h e a r e a b u t$ $a l s o o n h o w t h e a r e a i s d i s t r i b u t e d,$ $i . e . o n t h e s h a p e o f t h e a r e a, o r t h e$ $p o s i t i o n o f {a r e a}^{'} s c e n t e r o i d .$ $w h e n o n l y t h e a r e a i s g i v e n, t h e s t a t i c a l$ $m o m e n t i s n o t u n i q u e .$ $f o r e x a m p l e : t h e d o t t e d s h a p e h a s t h e$ $s a m e a r e a a s t h e s o l i d s h a p e, b u t t h e y$ $h a v e d i f f e r e n t s t a t i c a l m o m e n t a b o u t$ $y - a x i s, b e c a u s e t h e y h a v e d i f f e r e n t$ $p o s i t i o n o f c e n t r o i d .$ $t h i s i s t h e r e a s o n w h y \int_{0}^{1} x f (x) d x c a n$ $n o t b e u n i q u e d e t e r m i n e d w i t h t h e$ $g i v e n c o n d i t i o n s .$
Commented by mr W last updated on 04/Jan/19

Commented by mr W last updated on 04/Jan/19

$t h e r e i s n o u n i q u e s o l u t i o n .$
Commented by Otchere Abdullai last updated on 04/Jan/19

$p l e a s e w h i c h a p p d i d y o u u s e f o r y o u r$ $s k e t c h ? i w a n t t o d o w n l o a d$
Commented by mr W last updated on 04/Jan/19

$I u s e t h e a p p I N K r e d i b l e t o m a k e$ $s i m p l e s k e t c h e s .$
Commented by Otchere Abdullai last updated on 04/Jan/19

$t h a n k y o u s i r$
	Answered by pooja24 last updated on 04/Jan/19

	$x \underset{0}{\int^{1}} f (x) - \underset{0}{\int^{1}} \frac{d x}{d x} \int f (x) d x = {(3 x)}_{0}^{1} - \underset{0}{\int^{1}} 3 d x$ $= (3 - 0) - {(3 x)}_{0}^{1} = 3 - (3 - 0) = 3 - 3 = 0$ $n o t s u r e b u t t h i n k s o$
	Commented by prakash jain last updated on 04/Jan/19

	$x \underset{0}{\int^{1}} f (x) - \underset{0}{\int^{1}} \frac{d x}{d x} \int f (x) d x = {(3 x)}_{0}^{1} - \underset{0}{\int^{1}} 3 d x$ $You can not write$ $\int_{0}^{1} [\int f (x) d x] d x a s \int_{0}^{1} [\int_{0}^{1} f (x) d x] d x$ $You have evaluated internal integral$ $first .$
	Answered by MJS last updated on 04/Jan/19

	$if {it}^{'} s a polynome$ $f (x) = {a x}^{2} + b x + c$ $F (x) = \frac{a}{3} x^{3} + \frac{b}{2} x^{2} + c x$ $(1) c = 1$ $(2) a + b + 1 = 2 \Rightarrow a + b = 1$ $(3) \frac{a}{3} + \frac{b}{2} + 1 = 3 \Rightarrow 2 a + 3 b = 12$ $\Rightarrow b = 10, a = - 9$ $f (x) = - 9 x^{2} + 10 x + 1$ $\underset{0}{\int^{1}} x f (x) d x = \underset{0}{\int^{1}} (- 9 x^{3} + 10 x^{2} + x) d x =$ $= {[- \frac{9}{4} x^{4} + \frac{10}{3} x^{3} + \frac{1}{2} x^{2}]}_{0}^{1} = \frac{19}{12}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19

	$e x c e l l e n t s i r . . . d i f f e r e n t t h i n k t a n k . . .$
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