Definite integration and stuff

1. Apr 21, 2015

mooncrater

1. The problem statement, all variables and given/known data
The question says:
f (x)=x2+7x +∫0x(e-tf (x-t)dt.
Find f (x).

2. Relevant equations
None

3. The attempt at a solution
What I did is:
Consider the integral:
I=∫0x (e-tf (x-t)dt
We know that ∫abf (x)dx=∫abf (a+b-x)dc
So using it here:
1/ex0xetf (t)dt----(1)
Leaving the "1/ex part for now.
From the original equation f (t)=t2+7t +∫0te-t f (0)dt
And we can see that f (0)=0
So f (t)=t2+7t
Using (1) without 1/ex....
0xet(t2+7t)dt
Which can be calculated to ex(x2+5x -5)
Which is when divided by ex 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?

2. Apr 21, 2015

Dick

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.

Last edited: Apr 21, 2015
3. Apr 21, 2015

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.

Last edited: Apr 21, 2015
4. Apr 22, 2015

Raghav Gupta

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.

5. Apr 22, 2015

mooncrater

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?

6. Apr 22, 2015

Raghav Gupta

The first mistake is the bold part. You cannot treat 1/ex as constant.

7. Apr 22, 2015

mooncrater

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

8. Apr 22, 2015

Raghav Gupta

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

9. Apr 22, 2015

Raghav Gupta

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

10. Apr 22, 2015

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.

11. Apr 22, 2015

mooncrater

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?

12. Apr 22, 2015

Raghav Gupta

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.

13. Apr 22, 2015

Ray Vickson

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 $f(x) = x^2 + 7x + C$.
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.

14. Apr 22, 2015

Raghav Gupta

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

15. Apr 22, 2015

Ray Vickson

I thought I had cancelled 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.

16. Apr 22, 2015

Raghav Gupta

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

17. Apr 22, 2015

Ray Vickson

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.

18. Apr 22, 2015

Raghav Gupta

But you could have deleted the message.

19. Apr 22, 2015

Raghav Gupta

Is delete button removed recently from PF ?

20. Apr 22, 2015

Ray Vickson

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.