Find f(x) which satisfies this integral function

In summary: Isn't... isn't that a contradiction?No, it's not a contradiction. The derivative of the integral is zero because the function is constant at the boundaries of the interval.
  • #1
Helly123
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Homework Statement


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

Homework Equations

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?
 
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  • #2
Helly123 said:

Homework Statement


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

Homework Equations

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?
 
  • #3
Mark44 said:
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
 
  • #4
Helly123 said:
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.
 
  • #5
Helly123 said:

Homework Statement


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

Homework Equations

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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  • #6
Mark44 said:
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?
 
  • #7
Ray Vickson said:
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}##
 
  • #8
Helly123 said:
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##?
 
  • #9
Ray Vickson said:
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?
 
  • #10
Helly123 said:
##\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##?
 
  • #11
Ray Vickson said:
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?
 
  • #12
Helly123 said:
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##?
 
  • #13
Ray Vickson said:
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
 
  • #14
Helly123 said:
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##?
 
  • #15
Ray Vickson said:
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:
  • #16
Helly123 said:
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?
 
  • #17
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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  • #18
##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
 
  • #19
Helly123 said:
##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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1. How do I find the value of f(x) in an integral function?

To find the value of f(x) in an integral function, you need to evaluate the integral using the given limits of integration and any known information about the function. This will give you the definite integral, which represents the value of f(x) at a specific point or over a specific interval.

2. Can I solve for f(x) algebraically in an integral function?

In most cases, it is not possible to solve for f(x) algebraically in an integral function. This is because the integral represents the area under a curve, and there is no general formula for finding the antiderivative of a given function. However, you can approximate the value of f(x) using numerical methods or by using a computer program.

3. How do I determine the limits of integration for finding f(x) in an integral function?

The limits of integration for an integral function are typically given in the problem or can be determined from the context. They represent the range over which you are finding the value of f(x). For example, if the integral represents the area under a curve between x = 0 and x = 2, the limits of integration would be 0 and 2.

4. Can I use substitution or integration by parts to find f(x) in an integral function?

Yes, substitution and integration by parts are common techniques used to evaluate integrals and can be used to find f(x) in an integral function. However, these methods may not always result in a closed form solution, and numerical methods may be necessary.

5. Is there a specific method or formula for finding f(x) in an integral function?

No, there is no one specific method or formula for finding f(x) in an integral function. The method used will depend on the form of the integral and any known information about the function. Some common techniques include substitution, integration by parts, and numerical methods.

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