# Find $$f(x)$$

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 $$f(x)$$.

Note by Pratik Shastri
3 years, 8 months ago

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We 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 $$f(x)=C$$

- 3 years, 8 months ago

Comment deleted Oct 04, 2014

And why does it become meaningless?

- 3 years, 8 months ago

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

- 3 years, 8 months ago

Yeah..I too get a cubic that doesn't factorize

- 3 years, 8 months ago

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.

- 3 years, 8 months ago

Alright..I will change it to $$x^2$$..I don't think $$f(x)$$ can be expressed in terms of elementary functions..

- 3 years, 8 months ago

Hey Ronak, can you help me with this problem. https://brilliant.org/discussions/thread/iq-scores-jumble/

- 3 years, 8 months ago

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

- 3 years, 8 months ago