# Integral equation

1. Jan 21, 2009

### trebmling

Hello,

G(x)= INTEGRAL[F(t)dt]

lower integral limit: x-m,
upper integral limit: x,
m is a constant,
G(x) is a known function,
F(x) is unknown and should be found.

Thank you very much!

2. Jan 21, 2009

### NoMoreExams

Differentiate both sides?

You should get G'(x) = F(x) - F(x-m).

3. Jan 21, 2009

### trebmling

so easy? are you sure?

4. Jan 21, 2009

### NoMoreExams

Nope, I'm not in your class, I don't know what you are learning nor what the teacher is trying to show you with that exercise but I do like giving false leads so people would do bad on their homework.

5. Jan 21, 2009

### trebmling

I was thinking that there should be a more complex expression while differentiating th integral by a parameter x.

6. Jan 21, 2009

### NoMoreExams

Maybe it is, consult FTC

7. Jan 21, 2009

### trebmling

what's FTS?

8. Jan 21, 2009

### NoMoreExams

Not sure what FTS is but FTC is Fundamental Theorem of Calculus

9. Jan 21, 2009

### trebmling

In the Fundamental Theorem of Calculus, the f(x) under integral and the final F(x) are different functions, namely f(x)=F'(x). Therefore, in my case it does not help, since my function under the integral is unknown.

This equation is easy, but not that easy.

I am now trying to solve it by reformulating through series.

10. Jan 21, 2009

### NoMoreExams

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

Your G(x) is their F(x) therefore your G'(x) is their F'(x). Your F(t) is their f(t). The only difference is their integral goes from a to x and yours goes from x-m to x therefore you can generalize their version to

$$F(x) = \int_{\alpha(x)}^{\beta(x)} f(t) dt \Rightarrow F'(x) = f(\beta(x)) \cdot \beta'(x) - f(\alpha(x)) \cdot \alpha'(x)$$

11. Jan 21, 2009

### trebmling

nope, ...but may be you are right.

Last edited: Jan 21, 2009
12. Jan 21, 2009

### NoMoreExams

nope what?

13. Jan 21, 2009

### NoMoreExams

I wasn't claiming it's THAT easy to solve for F(x) it is not, in what I told you, I don't explicitly solve for F(x) I just give an exprsession that involves it and if your G(x) is "nice" enough you can "guess" what F(x) would have to be.

14. Jan 21, 2009

### trebmling

I have just derived another solution for my problem throught series, and it reads like this:

F(x)=G(x)- SUM(F(t)) {summation for t from x-m to x-1}

15. Jan 21, 2009

### trebmling

Your solution is probably right, and then these two solutions should be equivalent. If my solution through series is right.