21Sir, can u pl. post da crrct methd 4 integrtn by parts in "definite" by taking a general eg.
I too make mistakes many a times ther
The number of possible continuous f(x) defined in [0, 1] for which
I1 = \int_{0}^{1}{f(x)dx}= 1,
I2=\int_{0}^{1}{x.f(x)dx}=a
I3=\int_{0}^{1}{x^{2}.f(x)dx}=a^{2}
is ?
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24 Answers
21
341See one more instance of an application of this inequality:
http://targetiit.com/iit_jee_forum/posts/prove2_1532.html
341No, 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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341This 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
24arre sir.......meri galti to batao.....
aur apni site se hi bhaag rahe ho.....
62he was last running around trying to hide his face cos he could not solve this simple one!
1in I2 by applying ILATE(by parts).......
i got..x-1=a..
while in I3 by applyin by parts...
i got...x2-1=a2
that is x+1=a.........for a≠0...
so x+1=x-1=a.....which is not possible i guess,...
so ans is ZERO functions
24has anyone figured out whats wrong in my solution???????????[7][7]
24I 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]
1wel........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
62This is a very very good question..
Unfortunately I am unable to solve this one right nwo ankit..
I hope I can pull this off soon..
My guess is zero solutions!
62this is not correct akand.
you cant use definite integral like this..
indefinite willwork but not definite
1ok......
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
62can you show how you have applied integration by parts?
It is a good approach but I was unable to see how you got the values that you did!