Given that \[x^2\displaystyle\int_{0}^{x} (f(t))^3 \ dt=\left(\displaystyle\int_{0}^{x} f(t) \ dt \right)^3\] find all the possible functions \(f(x)\).

@Pratik Shastri
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With \({x}^{2}\) that would factorise easily and it can be solved, I checked the cubic that I got with x in wolfram alpha but it gave me weird solutions.

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## Comments

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TopNewestWe begin with writing :

\(g(x)=\int _{ 0 }^{ x }{ f(t)dt } \)

\(\Rightarrow {g}_{1}(x)=f(x)\) where \({g}_{1}(x)=\frac{dg(x)}{dx}\)

Our equation becomes \({x}^{2}\int _{ 0 }^{ x }{ { f(t) }^{ 3 }dt } = { g(x) }^{ 3 }\) (i)

Differentiating both sides with respect to x we get :

\(2x\int _{ 0 }^{ x }{ { f(t) }^{ 3 }dt }+{x}^{2} {({g}_{1}(x))}^{3}=3({g(x)}^{2})({g}^{'}(x))\) (ii)

Using (i) we write (ii) as :

\(2(\frac { { g(x })^{ 3 } }{ x }) + {x}^{2}{({g}_{1}(x))}^{3}=3({g(x)}^{2})({g}_{1}(x))\) (iii)

After some rearranging we get :

\({(x{g}_{1}(x))}^{3}-3{g(x)}^{2}(x{g}_{1}(x))+2{g(x)}^{3}=0\)

Dividing both sides by \({g(x)}^{3}\) we get :

\({(\frac { x{ g }_{ 1 }(x) }{ g(x) })}^{3}-3(\frac { x{ g }_{ 1 }(x) }{ g(x) }) +2 =0\)

Take \(\frac { x{ g }_{ 1 }(x) }{ g(x) } =y\) to get our equation as :

\({y}^{3}-3y+2=0\)

Factorising this we get :

\((y+2){(y-1)}^{2}=0\)

\(\Rightarrow y=-2,1\)

We will deal both these cases seperately :

Case-1,\(y=1\)

\(\frac { x{ g }_{ 1 }(x) }{ g(x) } =1\)

\(\Rightarrow \int { \frac { dg(x) }{ g(x) } } =\int { \frac { dx }{ x } } \)

\(\Rightarrow g(x)=Cx\),

Differentiating both sides we get :

\(f(x)=C \) (iv)

Case-2,\(y=-2\)

\(\frac { x{ g }_{ 1 }(x) }{ g(x) } =-2\)

\(\Rightarrow \int { \frac { dg(x) }{ g(x) } } =\int { \frac {-2dx }{ x } } \)

\(\Rightarrow g(x)=\frac{C}{{x}^{2}}\)

But since \(g(0)=0\) hence no value of \(C\) exists.

So the only function that exists is \(f(x)=C\)

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Comment deleted Oct 04, 2014

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And why does it become meaningless?

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Because it gave me an cubic that won't factorise easily.

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I did solve it while it was \(x^2\). The question I meant to ask contains an \(x\).

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