Definite integration and stuff

[mooncrater]

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

Homework Equations

None

The Attempt at a Solution

What I did is:
Consider the integral:
I=∫₀^x (e^-tf (x-t)dt
We know that ∫_a^bf (x)dx=∫_a^bf (a+b-x)dc
So using it here:
1/e^x∫₀^xe^tf (t)dt----(1)
Leaving the "1/e^x part for now.
From the original equation f (t)=t²+7t +∫₀^te^-t f (0)dt
And we can see that f (0)=0
So f (t)=t²+7t
Using (1) without 1/e^x...
∫₀^xe^t(t²+7t)dt
Which can be calculated to e^x(x²+5x -5)
Which is when divided by e^x becomes
x²+5x-5=I
Now adding I to the original equation of f (x)..
Thus f (x)= 2x²+12x -5x.
But in the solution they have done something else. So what is wrong with my solution to this question?

 

[Dick]

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

Homework Equations

None

The Attempt at a Solution

What I did is:
Consider the integral:
I=∫₀^x (e^-tf (x-t)dt
We know that ∫_a^bf (x)dx=∫_a^bf (a+b-x)dc
So using it here:
1/e^x∫₀^xe^tf (t)dt----(1)
Leaving the "1/e^x part for now.
From the original equation f (t)=t²+7t +∫₀^te^-t f (0)dt
And we can see that f (0)=0
So f (t)=t²+7t
Using (1) without 1/e^x...
∫₀^xe^t(t²+7t)dt
Which can be calculated to e^x(x²+5x -5)
Which is when divided by e^x becomes
x²+5x-5=I
Now adding I to the original equation of f (x)..
Thus f (x)= 2x²+12x -5x.
But in the solution they have done something else. So what is wrong with my solution to this question?

You are scrambling $x$ and $t$ up left and right. I don't see how you can conclude $f(t)=t^2+7t$ and then go on to derive that $f(x)=2x^2+12x-5$. Does that even make any sense?? I think you should look back and figure out where you went wrong yourself. There are really too many places to mention.

 

[certainly]

Interesting problem.
First off, you'll get quicker and more replies if you use LATEX to write your equations. It's not too hard to learn, see this. (I myself don't know much yet!)
Secondly, $$t$$ is a variable in the integral therefore by substituting for $$t$$ as a constant in the original equation, you can't write $$f(0)$$ in $$\int_0^x e^{-t}f(x-t)\ dt$$
Thirdly, in the solution that you saw was the answer recursive ?
Cheers.

 

[Raghav Gupta]

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

Homework Equations

None

The Attempt at a Solution

What I did is:
Consider the integral:
I=∫₀^x (e^-tf (x-t)dt
We know that ∫_a^bf (x)dx=∫_a^bf (a+b-x)dc
So using it here:
1/e^x∫₀^xe^tf (t)dt----(1)
Leaving the "1/e^x part for now.
From the original equation f (t)=t²+7t +∫₀^te^-t f (0)dt
And we can see that f (0)=0
So f (t)=t²+7t
Using (1) without 1/e^x...
∫₀^xe^t(t²+7t)dt
Which can be calculated to e^x(x²+5x -5)
Which is when divided by e^x becomes
x²+5x-5=I
Now adding I to the original equation of f (x)..
Thus f (x)= 2x²+12x -5x.
But in the solution they have done something else. So what is wrong with my solution to this question?

Others can continue to point where you went wrong. I am only telling you a fast approach.
I suggest you the Leibniz theorem. It's a handy tool whenever you see another variable involved in integral, like t here.

 

[mooncrater]

Interesting problem.
First off, you'll get quicker and more replies if you use LATEX to write your equations. It's not too hard to learn, see this. (I myself don't know much yet!)
Secondly, $$t$$ is a variable in the integral therefore by substituting for $$t$$ as a constant in the original equation, you can't write $$f(0)$$ in $$\int_0^x e^{-t}f(x-t)\ dt$$
Thirdly, in the solution that you saw was the answer recursive ?
Cheers.

Thanks...
And now I understand why I can't put $t$ in $f (x)$. So what I understand now is that this method is useful for nothing. I had one problem in my mind...
If I differentiate $f (x)$ using the Leibnitz rule, then I'll get:
$f'(x) = 2x+7+e^{-x}f (x-x)$
And since $f (0)=0${is it true? }
Therefore $$f'(x)=2x+7$$
And if we integrate it (sounds like I am again jumping back and forth):
$f (x)=x^2+7x +C$
So does that mean that the integral in the original equation is a constant?

 

[Raghav Gupta]

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

Homework Equations

None

The Attempt at a Solution

What I did is:
Consider the integral:
I=∫₀^x (e^-tf (x-t)dt
We know that ∫_a^bf (x)dx=∫_a^bf (a+b-x)dc
So using it here:
1/e^x∫₀^xe^tf (t)dt----(1)
Leaving the "1/e^x part for now.
From the original equation f (t)=t²+7t +∫₀^te^-t f (0)dt
And we can see that f (0)=0
So f (t)=t²+7t
Using (1) without 1/e^x...
∫₀^xe^t(t²+7t)dt
Which can be calculated to e^x(x²+5x -5)
Which is when divided by e^x becomes
x²+5x-5=I
Now adding I to the original equation of f (x)..
Thus f (x)= 2x²+12x -5x.
But in the solution they have done something else. So what is wrong with my solution to this question?

The first mistake is the bold part. You cannot treat 1/e^x as constant.

 

[mooncrater]

You cannot treat 1/ex as constant.

But Its an integral of $t$ therefore any thing else would be treated as a constant. Isn't it?

 

[Raghav Gupta]

Thanks...
And now I understand why I can't put $t$ in $f (x)$. So what I understand now is that this method is useful for nothing. I had one problem in my mind...
If I differentiate $f (x)$ using the Leibnitz rule, then I'll get:
$f'(x) = 2x+7+e^{-x}f (x-x)$
And since $f (0)=0${is it true? }
Therefore $$f'(x)=2x+7$$
And if we integrate it (sounds like I am again jumping back and forth):
$f (x)=x^2+7x +C$
So does that mean that the integral in the original equation is a constant?

Yeah it's a constant.
What is its value?

 

[Raghav Gupta]

But Its an integral of $t$ therefore any thing else would be treated as a constant. Isn't it?

No it is two variables. It's a parametric equation in terms of t.
That is why Leibniz theorem is handy here.

 

[Raghav Gupta]

For better understanding of this theorem see this thread.
It is bit similar to this question.
Certainly three of us were involved in that thread also.

 

[mooncrater]

No it is two variables

But an integral is of one variable only. That's why we use $dt$ here and not $dt$ as well as $dx$.
What I think is that the upper limit $x$ is just an $upper limit$. Though it's a $function$ of $x$ but the integral is of $t$. What do you think?

 

[Raghav Gupta]

But an integral is of one variable only. That's why we use $dt$ here and not $dt$ as well as $dx$.
What I think is that the upper limit $x$ is just an $upper limit$. Though it's a $function$ of $x$ but the integral is of $t$. What do you think?

Think of differential equations.
Think variable separation form. Correlate that all.
Here x can have any value, that is why it is a variable.
t is obvious as it is being integrated, upper limit therefore can have any value when x is changed.
That is why depending on x value integral would have different value.

 

[Ray Vickson]

Yeah it's a constant.
What is its value?

In the equation $f(x) = x^2 + 7x + \int_0^x e^{-t} f(x-t) \, dt$, what do you get by taking $x \to 0$?

Anyway, using a Laplace-transform approach I get a solution quite different from your supposed solution $f(x) = x^2 + 7x + C$.

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

Homework Equations

None

The Attempt at a Solution

What I did is:
Consider the integral:
I=∫₀^x (e^-tf (x-t)dt
We know that ∫_a^bf (x)dx=∫_a^bf (a+b-x)dc
So using it here:
1/e^x∫₀^xe^tf (t)dt----(1)
Leaving the "1/e^x part for now.
From the original equation f (t)=t²+7t +∫₀^te^-t f (0)dt
And we can see that f (0)=0
So f (t)=t²+7t
Using (1) without 1/e^x...
∫₀^xe^t(t²+7t)dt
Which can be calculated to e^x(x²+5x -5)
Which is when divided by e^x becomes
x²+5x-5=I
Now adding I to the original equation of f (x)..
Thus f (x)= 2x²+12x -5x.
But in the solution they have done something else. So what is wrong with my solution to this question?

If you know about Laplace transforms, using them would be the simplest way to go on this problem, because the last term $\int_0^x e^{-t} f(x-t) \, dt$ is the convolution of the two functions $f(x)$ and $g(x) = e^{-x}$ and so has transform equal to the product of the transforms of $f$ and $g$. You end up with a simple equation to determine the transform of $f$, and can then determine $f$ itself by transform inversion.

 

[Raghav Gupta]

In the equation $f(x) = x^2 + 7x + \int_0^x e^{-t} f(x-t) \, dt$, what do you get by taking $x \to 0$?

Anyway, using a Laplace-transform approach I get a solution quite different from your supposed solution $f(x) = x^2 + 7x + C$.

I know that what we get when we take x = 0 in equation. I was asking to OP.
Mooncrater can you confirm the answer?

 

[Ray Vickson]

I know that what we get when we take x = 0 in equation. I was asking to OP.
Mooncrater can you confirm the answer?

I thought I had canceled this response before it was posted, but somehow it went through anyway. It seems the PF document handling system is tricky and does not always produce desired results.

 

[Raghav Gupta]

I thought I had canceled this response before it was posted, but somehow it went through anyway. It seems the PF document handling system is tricky and does not always produce desired results.

It is still there. Are you talking of whole of post 13?

 

[Ray Vickson]

It is still there. Are you talking of whole of post 13?

Yes, but it is even more messed up than I realized. I started to reply to your post # 8, then decided to not post my response---I had changed my mind about responding. Later, I responded to the first post #1 and posted that. Somehow, the two things ended up in the same response to a totally different message from what I wanted. The response I never posted plus the response I did post both ended up in the same, wrong, message.

 

[Raghav Gupta]

But you could have deleted the message.

 

[Raghav Gupta]

Is delete button removed recently from PF ?

 

[Ray Vickson]

But you could have deleted the message.

How can I delete something if I don't know it is there? Besides, the other part of the final message was something I chose to keep.

Also: it used to be the case (I have not checked if it is still true) that once a message has been responded to, the original poster can no longer delete the message---that would allow for somebody to delete someone else's messages.

 

[Raghav Gupta]

Oh, I see the button is not available for homework forums.

Is the answer coming by Leibniz theorem wrong, f(x) = x² + 7x ?
What is the wrong part in post 5 ?

 

[Ray Vickson]

Oh, I see the button is not available for homework forums.

Is the answer coming by Leibniz theorem wrong, f(x) = x² + 7x ?
What is the wrong part in post 5 ?

The poster in #5 has mis-used Leibniz rule.
$$ \frac{d}{dx} \int_0^x e^{-t} f(x-t) \, dt = e^{-x} f(0) \; \Leftarrow \; \text{wrong!} \\
\frac{d}{dx} \int_0^x e^{-t} f(x-t) \, dt = e^{-x} f(0) + \int_0^x e^{-t} f'(x-t) \, dt \; \Leftarrow \; \text{correct}
$$

You can see right away that the function $f(x) = x^2 + 7x$ does NOT solve the equation
$$ f (x)=x^2+7x + \int_0^x e^{-t} f(x-t) \, dt, $$
because that would require that $f(x) = f(x) + \int_0^x e^{-t} f(x-t) \, dt,$ so would require that we have $\int_0^x e^{-t} f(x-t) \, dt = 0$ for all $x$---which is false.

 

[Mark44]

I thought I had canceled this response before it was posted, but somehow it went through anyway. It seems the PF document handling system is tricky and does not always produce desired results.

Ray, what might have happened is you pressed the +Quote button, which stores the text of the post in a buffer. If you wind up not actually replying to that particular post, the quoted text is still in the buffer, so when you reply to another post, I'm guessing that your reply included what was copied earlier.

 

[Raghav Gupta]

Isn't Leibniz rule
$$ \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) = f(h(x))h'(x) - f(g(x)) g'(x) $$ ?
So it should be
$$ \frac{d}{dx} \int_0^x e^{-t} f(x-t) = e^{-x}f(0) ? $$ as second term would be zero because differentiation of a constant which is the lower limit is zero

 

[certainly]

No the leibniz integral rule is:-
$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}f(x,t)\ dt=\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}\ dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}$$

 

[Raghav Gupta]

I think that df/dx should be δf/δx or a partial differentiation of f(x,t) with respect to x?

 

[certainly]

yup, was editing at the time.

 

[Raghav Gupta]

What is that, a term hiding in post 25 at right?

 

[certainly]

What are you talking about ?

 

[Ray Vickson]

Ray, what might have happened is you pressed the +Quote button, which stores the text of the post in a buffer. If you wind up not actually replying to that particular post, the quoted text is still in the buffer, so when you reply to another post, I'm guessing that your reply included what was copied earlier.

Thanks: that is exactly what happened.