Find Minimum

rohan, why dont you actually write out the steps.

Ok I(f) = \int_0^1 x^2 f(x) - x f^2(x) \ dx

Now \frac{\partial I}{\partial f} = 0

What next?

and one error in the last line it will be partial differentiation ..w.r.t f(x)

i am talking for each definite x that varies from 0 to 1 read my soln carefully

I think i have an easier method

if we maximise the given expression fixing x (i.e for a fixd value of x)
then we see that we get an expression which only depends on f(x) (as we have fixed x)

now if are is to be maximum if we can maximise this value then we will get f(x) in terms of x and thus the maximum possible value of this expression for any x

thus are will aso be maximum in that case

thus we diffrentiate it partially w.r.t x

we get f(x)=x/2 thus the ans as 1/16

This is the method. I flicked this one from goiit where i posted this solution about a month back.

Here, the stumbling block is f(x), because it can be just about any function. Then it strikes that you encounter a similar situation when you are looking for the global minimum or maximum of a quadratic

So there you isolate the constant and push the unknown x into a perfect square

Same thing is done here for f(x). Because we know how to integrate a polynomial in x. So we push f(x) into some perfect square and we are rid of it

Problems in JEE will use some crossover techniques like this. When you play around with the expression, your brain will trigger some response that we have encountered this situation before. follow such hunches and you will be through!

Let g(x)=x^2f(x)-xf^2(x). Complete the square in f(x) as follows:
g(x)=-x\left(f^2(x)-2\cdot f(x)\cdot \dfrac{x}{2}+\dfrac{x^2}{4}-\dfrac{x^2}{4}\right)
\Rightarrow\ g(x)=\dfrac{x^3}{4}-x\left(f(x)-\dfrac{x}{2}\right)^2\leq \dfrac{x^3}{4}
since 0 ≤ x ≤1.
Therefore,
\dfrac{x^3}{4}-g(x)\geq 0
\Rightarrow\ \int_0^1\left(\dfrac{x^3}{4}-g(x)\right)\mathrm{d}x\geq 0
\Rightarrow\ \int_0^1\dfrac{x^3}{4}\ \mathrm{d}x -\int_0^1g(x)\ \mathrm{d}x\geq 0
\Rightarrow\ \int_0^1g(x)\ \mathrm{d}x\leq \dfrac{1}{16}
Accordingly, the maximum value of \int_0^1 (x^2f(x)-xf^2(x))\ \mathrm{d}x is \dfrac{1}{16}. The equality is attained for f(x)=\dfrac{x}{2}.

you were answering a different question, thats all

have got a diff eqn..

let z= x²f(x) -xf²(x)

z' = x²f'(x) + 2xf(x) - f²(x) - x.2f(x).f'(x) = 0 [for maxima or minima]

f'(x) = [f²(x) - 2xf(x)]/[x² - 2xf(x)]

let f(x)=y

so we have: dy/dx = [y² - 2xy]/[x²-2xy] ... homogeneous..

am i doing anything worthful ?

RHS is a function of what variable?

You are differentiating w.r.t to what variable?

but i agree with you in the example

this cant be done the normal way then?

but since the limits are given to be constants

differentiating the entire thing would always give zero

how do you prove that?

If f(x) = 0, the integral is 0. If f(x) = 1, then it is -1/6. So, the integral is not constant