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Question-52161




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.
$${f}\left({x}\right)=\begin{cases}{\mathrm{7}{x}+\mathrm{1}}&{\mathrm{0}\leqslant{x}\leqslant\mathrm{1}/\mathrm{2}}\\{\mathrm{7}−\mathrm{5}{x}}&{\mathrm{1}/\mathrm{2}\leqslant{x}\leqslant\mathrm{1}}\end{cases} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} {f}\left({x}\right){dx}+\int_{\mathrm{1}\:/\mathrm{2}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\left[\frac{\mathrm{7}{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\right]_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} +\left[\mathrm{7}{x}−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \\ $$$$=\left(\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\mathrm{7}−\frac{\mathrm{5}}{\mathrm{2}}\right)−\left(\frac{\mathrm{7}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{7}+\mathrm{4}+\mathrm{56}−\mathrm{20}−\mathrm{28}+\mathrm{5}}{\mathrm{8}}=\frac{\mathrm{24}}{\mathrm{3}}=\mathrm{3} \\ $$$${f}\left({x}\right)\:{above}\:{satifies}\:{all}\:{conditons}. \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\mathrm{3} \\ $$$${xf}\left({x}\right)>\mathrm{0}\:{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}\:\mathrm{is}\:\mathrm{unlikely}\:\mathrm{to}\:\mathrm{be}\:\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{re}−\mathrm{check}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{my}\:\mathrm{workinhs} \\ $$$$\mathrm{again}. \\ $$
Commented by mr W last updated on 04/Jan/19
∫_0 ^1 f(x)dx represents the area. the  area is 3. but infinitely many shapes  can have the same area.  ∫_0 ^1 xf(x)dx represents the statical moment  of the area about y−axis. it depends  not only on the size of the area but  also on how the area is distributed,  i.e. on the shape of the area, or the  position of area′s centeroid.  when only the area is given, the statical  moment is not unique.  for example: the dotted shape has the  same area as the solid shape, but they  have different statical moment about  y−axis, because they have different  position of centroid.  this is the reason why ∫_0 ^1 xf(x)dx can  not be unique determined with the  given conditions.
$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:{represents}\:{the}\:{area}.\:{the} \\ $$$${area}\:{is}\:\mathrm{3}.\:{but}\:{infinitely}\:{many}\:{shapes} \\ $$$${can}\:{have}\:{the}\:{same}\:{area}. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}\:{represents}\:{the}\:{statical}\:{moment} \\ $$$${of}\:{the}\:{area}\:{about}\:{y}−{axis}.\:{it}\:{depends} \\ $$$${not}\:{only}\:{on}\:{the}\:{size}\:{of}\:{the}\:{area}\:{but} \\ $$$${also}\:{on}\:{how}\:{the}\:{area}\:{is}\:{distributed}, \\ $$$${i}.{e}.\:{on}\:{the}\:{shape}\:{of}\:{the}\:{area},\:{or}\:{the} \\ $$$${position}\:{of}\:{area}'{s}\:{centeroid}. \\ $$$${when}\:{only}\:{the}\:{area}\:{is}\:{given},\:{the}\:{statical} \\ $$$${moment}\:{is}\:{not}\:{unique}. \\ $$$${for}\:{example}:\:{the}\:{dotted}\:{shape}\:{has}\:{the} \\ $$$${same}\:{area}\:{as}\:{the}\:{solid}\:{shape},\:{but}\:{they} \\ $$$${have}\:{different}\:{statical}\:{moment}\:{about} \\ $$$${y}−{axis},\:{because}\:{they}\:{have}\:{different} \\ $$$${position}\:{of}\:{centroid}. \\ $$$${this}\:{is}\:{the}\:{reason}\:{why}\:\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}\:{can} \\ $$$${not}\:{be}\:{unique}\:{determined}\:{with}\:{the} \\ $$$${given}\:{conditions}. \\ $$
Commented by mr W last updated on 04/Jan/19
Commented by mr W last updated on 04/Jan/19
there is no unique solution.
$${there}\:{is}\:{no}\:{unique}\:{solution}. \\ $$
Commented by Otchere Abdullai last updated on 04/Jan/19
please which app did you use for your  sketch?  i want to download
$${please}\:{which}\:{app}\:{did}\:{you}\:{use}\:{for}\:{your} \\ $$$${sketch}?\:\:{i}\:{want}\:{to}\:{download} \\ $$
Commented by mr W last updated on 04/Jan/19
I use the app INKredible to make  simple sketches.
$${I}\:{use}\:{the}\:{app}\:{INKredible}\:{to}\:{make} \\ $$$${simple}\:{sketches}. \\ $$
Commented by Otchere Abdullai last updated on 04/Jan/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by pooja24 last updated on 04/Jan/19
x∫_0 ^1 f(x)−∫_0 ^1 (dx/dx)∫f(x)dx=(3x)_0 ^1 −∫_0 ^1 3 dx  =(3−0)−(3x)_0 ^1 =3−(3−0)=3−3=0  not sure but think so
$${x}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right)−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{dx}}\int{f}\left({x}\right){dx}=\left(\mathrm{3}{x}\right)_{\mathrm{0}} ^{\mathrm{1}} −\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{3}\:{dx} \\ $$$$=\left(\mathrm{3}−\mathrm{0}\right)−\left(\mathrm{3}{x}\right)_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{3}−\left(\mathrm{3}−\mathrm{0}\right)=\mathrm{3}−\mathrm{3}=\mathrm{0} \\ $$$${not}\:{sure}\:{but}\:{think}\:{so} \\ $$
Commented by prakash jain last updated on 04/Jan/19
x∫_0 ^1 f(x)−∫_0 ^1 (dx/dx)∫f(x)dx=(3x)_0 ^1 −∫_0 ^1 3 dx  You can not write  ∫_0 ^1 [∫f(x)dx]dx as ∫_0 ^1 [∫_0 ^1 f(x)dx]dx  You have evaluated internal integral  first.
$${x}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right)−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{dx}}\int{f}\left({x}\right){dx}=\left(\mathrm{3}{x}\right)_{\mathrm{0}} ^{\mathrm{1}} −\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{3}\:{dx} \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{not}\:\mathrm{write} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int{f}\left({x}\right){dx}\right]{dx}\:{as}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\right]{dx} \\ $$$$\mathrm{You}\:\mathrm{have}\:\mathrm{evaluated}\:\mathrm{internal}\:\mathrm{integral} \\ $$$$\mathrm{first}. \\ $$
Answered by MJS last updated on 04/Jan/19
if it′s a polynome  f(x)=ax^2 +bx+c  F(x)=(a/3)x^3 +(b/2)x^2 +cx  (1)   c=1  (2)   a+b+1=2 ⇒ a+b=1  (3)   (a/3)+(b/2)+1=3 ⇒ 2a+3b=12  ⇒ b=10, a=−9  f(x)=−9x^2 +10x+1  ∫_0 ^1 xf(x)dx=∫_0 ^1 (−9x^3 +10x^2 +x)dx=  =[−(9/4)x^4 +((10)/3)x^3 +(1/2)x^2 ]_0 ^1 =((19)/(12))
$$\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{polynome} \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${F}\left({x}\right)=\frac{{a}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{{b}}{\mathrm{2}}{x}^{\mathrm{2}} +{cx} \\ $$$$\left(\mathrm{1}\right)\:\:\:{c}=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\:\:{a}+{b}+\mathrm{1}=\mathrm{2}\:\Rightarrow\:{a}+{b}=\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\:\:\frac{{a}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}+\mathrm{1}=\mathrm{3}\:\Rightarrow\:\mathrm{2}{a}+\mathrm{3}{b}=\mathrm{12} \\ $$$$\Rightarrow\:{b}=\mathrm{10},\:{a}=−\mathrm{9} \\ $$$${f}\left({x}\right)=−\mathrm{9}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{1} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xf}\left({x}\right){dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(−\mathrm{9}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} +{x}\right){dx}= \\ $$$$=\left[−\frac{\mathrm{9}}{\mathrm{4}}{x}^{\mathrm{4}} +\frac{\mathrm{10}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{19}}{\mathrm{12}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19
excellent sir...different think tank...
$${excellent}\:{sir}…{different}\:{think}\:{tank}… \\ $$

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