integration

See one more instance of an application of this inequality:

http://targetiit.com/iit_jee_forum/posts/prove2_1532.html

No, no, there is a straightforward way too (which is not the book solution). I thought this is a good place to introduce this useful inequality

Just notice that a^2 I_1 -2a I_2 + I_3 = 0

But, this implies that \int_0^a f(x) (x-a)^2 dx = 0

Since both f(x) and (x-a)² are non-negative and not identicall equal to zero in the given interval, this is not possible.

Hence no such functions exist.

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This is a known problem. But the condition that is missing in the problem statement given here, is that f(x) is non-negative.

You will need the following result:

Schwarz Bunyakovsky Inequality -

\left( \int_0^a f(x) g(x) dx \right)^2 \le \int_0^a f^2(x) dx \int_0^a g^2 (x) dx

I will give an easy proof here that I read in an interesting book 'Cauchy Schwarz Master Class' [this inequality resembles the Cauchy Schwarz Inequality]

Proof:

For some real t, we have \int_0^a (tf(x)- g(x))^2 dx \ge 0

Now,

\int_0^a (tf(x)- g(x))^2 dx = \int_0^a f^2(x) dx t^2 - 2t \int_0^a f(x) dx \int_0^a g(x) dx + \int_0^a g^2(x) dx

which is obviously a quadratic in t. For this to assume non-negative

values we must have D \le 0 where D = 4 \left[\left( \int_0^a f(x) dx \int_0^a g(x) dx \right)^2 - \int_0^a f^2(x) dx \int_0^a g^2(x) dx \right] and you can see that the given inequality immediately follows.

Its important to note when equality holds. Its when f(x) = t g(x) for x \in [0,a]

Now, to the problem itself.

We have \left( \int_0^a x f(x) dx \right)^2 \le \int_0^a \left(\sqrt{f(x)} \right)^2 dx \int_0^a \left(x \sqrt{f(x)} \right)^2 dx

But from the given conditions we see that equality holds here.

This is possible iff \sqrt{f(x)} = \lambda x \sqrt{f(x)} which is absurd as f cannot be identically zero from the 1st integral.

Hence no such functions exist

arre sir.......meri galti to batao.....
aur apni site se hi bhaag rahe ho.....

"this simple one!"????????

[11]

where is sir?????????

I tried to do it this way..............which obviously doesnt lead anywhere....
I₁=F(x)]₀¹ where F(x) is anti derivative of f(x)
=>I₁=F(1)- F(0)=1 ---------------------(1)

For I₂ taking x as first function and integrating........
1
x.F(x)]₀¹ - ∫ F(x)dx
0
=>I₂=F(1)-G(1)+G(0)=a ----------(2) where G(x) is anti derivative of F(x)

I₃=F(1)-2[G(1)-H(1)+H(0)] =a²------(3) where H(x) is anti derivative of G(x)

[2][2]

Answer is 1.

wel........lets do by examples.............
let f(x) be 1..
so
I2=∫xdx=x²/2=1/2=a
I3=∫x²dx=x³/3=1/3=a²

this is highly impossible (a=1/2....a²=1/3)

so zero solutions

Options :
0,1,2 ,infinity

ok...........thnx bhaiyya....

ok......
I2=x₀∫¹f(x)dx-₀∫¹1.(∫f(x)dx)dx
=x-x]_o¹
= x-1-0
=x-1
= a (given)
any mistake bhaiyya???? similarly for I3