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Question Number 56147 by afachri last updated on 11/Mar/19
if ∫_( 1) ^( 2) f(x) dx = (√( 2 )), then ∫_( 1) ^( 4) (1/( (√( x )) )) f(x) dx is ?? please help me Sir. I′ve been trying this for 2 days and getting stuck.
$$\mathrm{if}\:\underset{\:\:\mathrm{1}} {\overset{\:\:\mathrm{2}} {\int}}\:{f}\left({x}\right)\:{dx}\:=\:\sqrt{\:\mathrm{2}\:},\:\mathrm{then}\:\underset{\:\:\mathrm{1}} {\overset{\:\:\mathrm{4}} {\int}}\:\frac{\mathrm{1}}{\:\sqrt{\:{x}\:}\:}\:{f}\left({x}\right)\:{dx} \\ $$$$\:\mathrm{is}\:?? \\ $$$$\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{Sir}.\:\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{trying} \\ $$$$\:\:\mathrm{this}\:\mathrm{for}\:\mathrm{2}\:\mathrm{days}\:\mathrm{and}\:\mathrm{getting}\:\mathrm{stuck}. \\ $$$$ \\ $$$$ \\ $$
Commented by 121194 last updated on 11/Mar/19
f(x)=(√2) ∫_1 ^2 f(x)dx=(√2) ∫_1 ^4 ((f(x))/( (√x)))dx=2(√2) ∫_1 ^2 (√x)dx=(2/3)(2(√2)−1) f(x)=((3(√(2x)))/(2(2(√2)−1))) ∫_1 ^2 f(x)dx=(√2) ∫_1 ^4 ((f(x))/( (√x)))dx=((9(√2))/(2(2(√2)−1)))=((9(4+(√2)))/(14)) ∫_1 ^4 ((f(x))/( (√x)))dx u^2 =x⇒u=(√x);du=(dx/(2(√x)))⇔2du=(dx/( (√x))) x=1⇒u=1 x=4⇒u=2 ∫_1 ^4 f(x)(dx/( (√x)))=2∫_1 ^2 f(u^2 )du
$${f}\left({x}\right)=\sqrt{\mathrm{2}} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\sqrt{\mathrm{2}} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\sqrt{{x}}{dx}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{2}{x}}}{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\sqrt{\mathrm{2}} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)}=\frac{\mathrm{9}\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)}{\mathrm{14}} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx} \\ $$$${u}^{\mathrm{2}} ={x}\Rightarrow{u}=\sqrt{{x}};{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}}\Leftrightarrow\mathrm{2}{du}=\frac{{dx}}{\:\sqrt{{x}}} \\ $$$${x}=\mathrm{1}\Rightarrow{u}=\mathrm{1} \\ $$$${x}=\mathrm{4}\Rightarrow{u}=\mathrm{2} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right)\frac{{dx}}{\:\sqrt{{x}}}=\mathrm{2}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{f}\left({u}^{\mathrm{2}} \right){du} \\ $$
Commented by 121194 last updated on 11/Mar/19
are you sure this question is right?
$$\mathrm{are}\:\mathrm{you}\:\mathrm{sure}\:\mathrm{this}\:\mathrm{question}\:\mathrm{is}\:\mathrm{right}? \\ $$
Commented by afachri last updated on 11/Mar/19
yes Sir, that′s the question is Sir
$$\mathrm{yes}\:\mathrm{Sir},\:\mathrm{that}'{s}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{S}{ir} \\ $$
Commented by afachri last updated on 11/Mar/19
pardon me Sir. I don′t undersrand your solution. Please rewrite it clearly Sir.
$${pardon}\:{me}\:{Sir}.\:{I}\:{don}'{t}\:{undersrand} \\ $$$${your}\:{solution}.\:{Please}\:{rewrite}\:{it} \\ $$$${clearly}\:{Sir}. \\ $$
Commented by 121194 last updated on 11/Mar/19
wierd, i don′t think that condition is enought to determinate the other. because unless i made a typo, there2 diferet function such ∫_1 ^2 f(x)dx=(√2), but ∫_1 ^4 ((f(x))/( (√x)))dx gives diferent values also given g:[1,4]→R integable on that domain, you can always build a solution to it with f(x)=(((√2)g(x))/(∫_1 ^2 g(x)dx));∫_1 ^2 g(x)dx≠0
$$\mathrm{wierd},\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}\:\mathrm{condition}\:\mathrm{is}\:\mathrm{enought}\:\mathrm{to} \\ $$$$\mathrm{determinate}\:\mathrm{the}\:\mathrm{other}. \\ $$$$\mathrm{because}\:\mathrm{unless}\:\mathrm{i}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{there2}\:\mathrm{diferet}\:\mathrm{function} \\ $$$$\mathrm{such}\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\sqrt{\mathrm{2}},\:\mathrm{but}\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}\:\mathrm{gives}\:\mathrm{diferent}\:\mathrm{values} \\ $$$$\mathrm{also}\:\mathrm{given}\:{g}:\left[\mathrm{1},\mathrm{4}\right]\rightarrow\mathbb{R}\:\mathrm{integable}\:\mathrm{on}\:\mathrm{that}\:\mathrm{domain}, \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{always}\:\mathrm{build}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{it}\:\mathrm{with} \\ $$$${f}\left({x}\right)=\frac{\sqrt{\mathrm{2}}{g}\left({x}\right)}{\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{g}\left({x}\right){dx}};\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{g}\left({x}\right){dx}\neq\mathrm{0} \\ $$
Commented by prakash jain last updated on 12/Mar/19
If we modify the question∫_1 ^4 ((f((√)x))/( (√x)))dx x=u^2 dx=2udu ∫_1 ^4 ((f((√)x))/( (√x)))dx=∫_1 ^2 2f(u)du
$$\mathrm{If}\:\mathrm{we}\:\mathrm{modify}\:\mathrm{the}\:\mathrm{question}\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{f}\left(\sqrt{}{x}\right)}{\:\sqrt{{x}}}{dx} \\ $$$${x}={u}^{\mathrm{2}} \\ $$$${dx}=\mathrm{2}{udu} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{f}\left(\sqrt{}{x}\right)}{\:\sqrt{{x}}}{dx}=\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{2}{f}\left({u}\right){du} \\ $$
Answered by MJS last updated on 11/Mar/19
if we don′t know the nature of f(x) we can′t solve. the number of functions satisfying the 1^(st) condition is ∞
$$\mathrm{if}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{we}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{solve}.\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{functions}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{1}^{\mathrm{st}} \:\mathrm{condition}\:\mathrm{is}\:\infty \\ $$
Commented by MJS last updated on 11/Mar/19
even if f(x) is linear it′s not determined: f(x)=ax+b ∫_1 ^2 f(x)dx=((3a)/2)+b=(√2) ⇒ b=(√2)−((3a)/2) f(x)=ax+(√2)−((3a)/2); a∈R ∫_1 ^4 ((f(x))/( (√x)))dx=((5a)/3)+2(√2); a∈R if b=0 ⇒ a=((2(√2))/3) ⇒ ∫_1 ^4 ((f(x))/( (√x)))dx=((28(√2))/9) if a=0 ⇒ b=(√2) ⇒ ∫_1 ^4 ((f(x))/( (√x)))dx=2(√2) ...
$$\mathrm{even}\:\mathrm{if}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{linear}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{determined}: \\ $$$${f}\left({x}\right)={ax}+{b} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\frac{\mathrm{3}{a}}{\mathrm{2}}+{b}=\sqrt{\mathrm{2}}\:\Rightarrow\:{b}=\sqrt{\mathrm{2}}−\frac{\mathrm{3}{a}}{\mathrm{2}} \\ $$$${f}\left({x}\right)={ax}+\sqrt{\mathrm{2}}−\frac{\mathrm{3}{a}}{\mathrm{2}};\:{a}\in\mathbb{R} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\frac{\mathrm{5}{a}}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}};\:{a}\in\mathbb{R} \\ $$$$ \\ $$$$\mathrm{if}\:{b}=\mathrm{0}\:\Rightarrow\:{a}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\:\Rightarrow\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\frac{\mathrm{28}\sqrt{\mathrm{2}}}{\mathrm{9}} \\ $$$$\mathrm{if}\:{a}=\mathrm{0}\:\Rightarrow\:{b}=\sqrt{\mathrm{2}}\:\Rightarrow\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\frac{{f}\left({x}\right)}{\:\sqrt{{x}}}{dx}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$… \\ $$
Commented by afachri last updated on 11/Mar/19
thank you for all of You Sir. that was i thinking before. how to or is that any possible to do to find the real f(x)
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{all}\:\mathrm{of}\:{You}\:\mathrm{S}{ir}. \\ $$$$\mathrm{that}\:\mathrm{was}\:\mathrm{i}\:\mathrm{thinking}\:\mathrm{before}.\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{or}\:\mathrm{is}\:\mathrm{that}\:\mathrm{any}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{do}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{real}\:{f}\left({x}\right)\: \\ $$

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