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Find f(x) which satisfies this integral function

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1. Homework Statement
find f(x) which satisfies f(x) = x + ##\frac{1}{\pi}## ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

2. Homework Equations


3. The Attempt at a Solution
to solve f(x), I have to solve the integral which contains f(t). And f(t) is the f(x) with variable t? if yes, I will get integral within integral and within integral. Or what... should I do?
 
32,838
4,562
1. Homework Statement
find f(x) which satisfies f(x) = x + ##\frac{1}{\pi}## ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

2. Homework Equations


3. The Attempt at a Solution
to solve f(x), I have to solve the integral which contains f(t). And f(t) is the f(x) with variable t? if yes, I will get integral within integral and within integral. Or what... should I do?
Yes, f is the same in both places. As written, I don't have any suggestions, but are you sure that the integral is exactly as you wrote it? In particular, are you sure that the upper limit of integration is ##\pi##, and not x?
 
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but are you sure that the integral is exactly as you wrote it?
yes, i've confirmed it

In particular, are you sure that the upper limit of integration is ##\pi##, and not x?
yes it's ##\pi## , if it were x, then integral f(t) equals to f(x)
with the integral sin..

or kind like that.. i'm still trying
 
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yes it's ##\pi## , if it were x, then integral f(t) equals to f(x)
with the integral sin..
No, that doesn't make any sense.
Helly123 said:
or kind like that.. i'm still trying
So here's your equation:
##f(x) = x + \frac{1}{\pi}## ##\int_0^{\pi} f(t) \sin^2{t} \ dt##
What happens if you differentiate both sides? Keep in mind that the definite integral on the right side is a number, so its derivative is zero.
 

Ray Vickson

Science Advisor
Homework Helper
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1. Homework Statement
find f(x) which satisfies f(x) = x + ##\frac{1}{\pi}## ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

2. Homework Equations


3. The Attempt at a Solution
to solve f(x), I have to solve the integral which contains f(t). And f(t) is the f(x) with variable t? if yes, I will get integral within integral and within integral. Or what... should I do?
So, you have ##f(x) = x\, + \, c/ \pi,## where the constant ##c## happens to be given by ##c =\int_0^{\pi} f(t) \sin^2(t) \, dt##. You ought to be able to develop a simple equation for determining ##c##.
 
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No, that doesn't make any sense.

So here's your equation:
##f(x) = x + \frac{1}{\pi}## ##\int_0^{\pi} f(t) \sin^2{t} \ dt##
What happens if you differentiate both sides? Keep in mind that the definite integral on the right side is a number, so its derivative is zero.
hmm, the integral contains x, still can be zero?
 
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So, you have ##f(x) = x\, + \, c/ \pi,## where the constant ##c## happens to be given by ##c =\int_0^{\pi} f(t) \sin^2(t) \, dt##. You ought to be able to develop a simple equation for determining ##c##.
do you mean :

y = mx + c
y = x + ##\frac{\int_{0}^{\pi} f(t) \sin^2t \ d(t)}{\pi}##
 

Ray Vickson

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do you mean :

y = mx + c
y = x + ##\frac{\int_{0}^{\pi} f(t) \sin^2t \ d(t)}{\pi}##
No, I mean exactly what I wrote. You have ##f(x) = x + c/\pi## for some constant ##c## (so, of course, ##f(t) = t + c/ \pi## or ##f(w) = w + c/ \pi##). Can you do the integral ##\int_0^{\pi} f(t) \sin^2(t) \, dt##?
 
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Can you do the integral ##\int_0^{\pi} f(t) \sin^2(t) \, dt##?
##\int_{0}^{\pi} f(t) \sin^2 t \ d(t)##
= ##\int_{0}^{\pi} f(t) \frac{1}{2}(1 - \cos{2t}) \ d(t)##
= ## f(t) (\frac{1}{2}t - \frac{1}{2}2sin2t) |_{\pi}^{0} - \int_{0}^{\pi} \frac{1}{2}t - sin2t \ d(f(t)) \ d(t)##

what i'm not sure about is the derivative of f(t) which is the integral
f(t) = x + ' integral '
d(f(t))/d(t) = 1 + 0 = 1
is the derivative of the integral ( f( t ) ) is 0?
 

Ray Vickson

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##\int_{0}^{\pi} f(t) \sin^2 t \ d(t)##
= ##\int_{0}^{\pi} f(t) \frac{1}{2}(1 - \cos{2t}) \ d(t)##
= ## f(t) (\frac{1}{2}t - \frac{1}{2}2sin2t) |_{\pi}^{0} - \int_{0}^{\pi} \frac{1}{2}t - sin2t \ d(f(t)) \ d(t)##

what i'm not sure about is the derivative of f(t) which is the integral
f(t) = x + ' integral '
d(f(t))/d(t) = 1 + 0 = 1
is the derivative of the integral ( f( t ) ) is 0?
If you know that ##f(t) = t + c/\pi##, what is preventing you from computing ##\int_0^{\pi} f(t) \sin^2(t) \, dt##?
 
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If you know that ##f(t) = t + c/\pi##, what is preventing you from computing ##\int_0^{\pi} f(t) \sin^2(t) \, dt##?
By taking the derivative of both sides
Resulting
d f(t)/dt = 1
Isn't it?
 

Ray Vickson

Science Advisor
Homework Helper
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By taking the derivative of both sides
Resulting
d f(t)/dt = 1
Isn't it?
I will ask you just one more time:
Can you do the integral ##\int_0^{\pi} t \sin^2(t) \, dt?## Can you do the integral ##\int_0^{\pi} k \sin^2(t) \, dt## for any constant ##k##? If you can, what is preventing you from computing ##\int_0^{\pi} f(t) \, dt## for a function ##f## of the form ##f(t) = t + k##?
 
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I will ask you just one more time:
Can you do the integral ##\int_0^{\pi} t \sin^2(t) \, dt?## Can you do the integral ##\int_0^{\pi} k \sin^2(t) \, dt## for any constant ##k##? If you can, what is preventing you from computing ##\int_0^{\pi} f(t) \, dt## for a function ##f## of the form ##f(t) = t + k##?
I don't know exactly to integrate it...
I got ##\frac{{\pi}^2}{4}## as the answer.. but I'm not sure
 

Ray Vickson

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I don't know exactly to integrate it...
I got ##\frac{{\pi}^2}{4}## as the answer.. but I'm not sure
What happened to the term involving ##k##?
 
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What happened to the term involving ##k##?
The integral of k is ##\pi^2##/4
The f(x) = t + k/##\pi##
f(x) = x + ##\pi##/4
But
The right answer is f(x) = x + ##\pi##/2

Set f(t) = u and d(u) = 1
##\frac{1}{2}(1-\cos2t) ##= d(v) and v = ## \frac{1}{2} t - \frac{\sin2t}{2} . 2##

##\int u.d(v) = u.v - \int v. d(u) ##

##\int_{0}^{\pi} f(t) {\sin}^2t d(t)##
= ##\int_{0}^{\pi} f(t) \frac{1}{2}(1-\cos2t)##
= ##f(t) \frac{1}{2} t - \frac{\sin2t}{2} . 2 ]_{0}^{\pi} - \int_{0}^{\pi}\frac{1}{2}t - \frac{\sin_{2t}}{2} . 1##
 
Last edited:
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The integral of k is ##\pi^2##/4
##\int k dt = kt + C##
Helly123 said:
The f(x) = t + k/##\pi##
f(x) = x + ##\pi##/4
Which one is f(x)?
Also, f(x) should involve x, not t.
Helly123 said:
But
The right answer is f(x) = x + ##\pi##/2

Set f(t) = u and d(u) = 1
##\frac{1}{2}(1-\cos2t) ##= d(v) and v = ## \frac{1}{2} t - \frac{\sin2t}{2} . 2##

##\int u.d(v) = u.v - \int v. d(u) ##

##\int_{0}^{\pi} f(t) {\sin}^2t d(t)##
= ##\int_{0}^{\pi} f(t) \frac{1}{2}(1-\cos2t)##
= ##f(t) \frac{1}{2} t - \frac{\sin2t}{2} . 2 ]_{0}^{\pi} - \int_{0}^{\pi}\frac{1}{2}t - \frac{\sin_{2t}}{2} . 1##
Why didn't you substitute for f(t) in the integral?
 
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Poster has been reminded not to post solutions to homework questions
Hello Helly123, I think you are having some trouble to understand what Mark44 said, so I'll try to explain it in other words

Can you understand that ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## is contstant and does not depend on any variable?
So this wold give ##f(x) = x+c## for a constant c that is equal to the integral above divided by ##\pi##

Now you have to integrate ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t) = \int_{0}^{\pi} (t+c) \sin^2{t} \ d(t) = \pi (2c + \pi)/4##

And then you find c = ##\pi/2##
 
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##f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## = constant
I just get that it is constant because its variable doesn't depend on any x variable in f(x). because its variable is t . Is it true?

set ## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## as c for example
so, ##f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## become : ##f(x) = x + \frac{c}{\pi} ##

solve ## c = \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

##f(x) = x + \frac{c}{\pi} ##
##f(t) = t + \frac{c}{\pi} ##
c =## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##
c = ##\int_{0}^{\pi} (t + \frac{c}{\pi} ) \sin^2{t} \ d(t)##
= ##\int_{0}^{\pi} (t + \frac{c}{\pi} ) \frac{1}{2}(1 - \cos 2t) \ d(t)##
= ##\int_{0}^{\pi} (t \frac{1}{2}(1 - \cos 2t) + \frac{c}{\pi} \frac{1}{2}(1 - \cos 2t) \ d(t) )##
= ## \frac{1}{2} [ \int_{0}^{\pi}t \ d(t) - \int_{0}^{\pi} \cos 2t \ d(t) ] + \frac{c}{\pi} \frac{1}{2}[ \int_{0}^{\pi} \ d(t) - \int_{0}^{\pi} \cos 2t \ d(t) ]##


c = ## \frac{ \pi^2}{4} + \frac{c}{2} ##
##\frac{c}{2} ## = ## \frac{ \pi^2}{4}##
c = ## \frac{ \pi^2}{2}##


##f(x) = x + \frac{c}{\pi} ##
##f(x) = x + \frac{\frac{ \pi^2}{2}}{\pi} ##
##f(x) = x + \frac{ \pi}{2}##

Thanks for the explanation before
CMIIW
 
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##f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## = constant
I just get that it is constant because its variable doesn't depend on any x variable in f(x). because its variable is t . Is it true?

set ## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## as c for example
so, ##f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## become : ##f(x) = x + \frac{c}{\pi} ##

solve ## c = \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##

##f(x) = x + \frac{c}{\pi} ##
##f(t) = t + \frac{c}{\pi} ##
c =## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##
c = ##\int_{0}^{\pi} (t + \frac{c}{\pi} ) \sin^2{t} \ d(t)##
= ##\int_{0}^{\pi} (t + \frac{c}{\pi} ) \frac{1}{2}(1 - \cos 2t) \ d(t)##
= ##\int_{0}^{\pi} (t \frac{1}{2}(1 - \cos 2t) + \frac{c}{\pi} \frac{1}{2}(1 - \cos 2t) \ d(t) )##
= ## \frac{1}{2} [ \int_{0}^{\pi}t \ d(t) - \int_{0}^{\pi} \cos 2t \ d(t) ] + \frac{c}{\pi} \frac{1}{2}[ \int_{0}^{\pi} \ d(t) - \int_{0}^{\pi} \cos 2t \ d(t) ]##


c = ## \frac{ \pi^2}{4} + \frac{c}{2} ##
##\frac{c}{2} ## = ## \frac{ \pi^2}{4}##
c = ## \frac{ \pi^2}{2}##


##f(x) = x + \frac{c}{\pi} ##
##f(x) = x + \frac{\frac{ \pi^2}{2}}{\pi} ##
##f(x) = x + \frac{ \pi}{2}##

Thanks for the explanation before
CMIIW
Actually it does not depend on t either. That’s because the integral has the limits defined!
The function ##f(t)sin^2(t)## does depend on t, but the integral of that from zero to ##\pi## is a number!
 

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