Find f(x) which satisfies this integral function

[Helly123]

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?

 

[Mark44]

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?

 

[Helly123]

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

 

[Mark44]

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.

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]

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 $c$.

 

[Helly123]

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?

 

[Helly123]

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]

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 $\int_0^{\pi} f(t) \sin^2(t) \, dt$?

 

[Helly123]

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]

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

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 $\int_0^{\pi} f(t) \sin^2(t) \, dt$?

 

[Helly123]

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]

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 $f(t) = t + k$?

 

[Helly123]

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]

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 $k$?

 

[Helly123]

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 = $\frac{1}{2} t - \frac{\sin2t}{2} . 2$

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

$\int_{0}^{\pi} f(t) {\sin}^2t d(t)$
= $\int_{0}^{\pi} f(t) \frac{1}{2}(1-\cos2t)$
= $f(t) \frac{1}{2} t - \frac{\sin2t}{2} . 2 ]_{0}^{\pi} - \int_{0}^{\pi}\frac{1}{2}t - \frac{\sin_{2t}}{2} . 1$

 

[Mark44]

The integral of k is $\pi^2$/4

$\int k dt = kt + C$

The f(x) = t + k/$\pi$
f(x) = x + $\pi$/4

Which one is f(x)?
Also, f(x) should involve x, not t.

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?

 

[jaumzaum]

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 = $\pi/2$

 

[Helly123]

$f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)$

$\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)$ = 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

 

[jaumzaum]

$f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)$

$\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)$ = 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!