Solve Integral Equation | G(x) and F(x)

[trebmling]

Hello,

Could you please help me to solve this equation:

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!

 

[NoMoreExams]

Differentiate both sides?

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

 

[trebmling]

Differentiate both sides?

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

so easy? are you sure?

 

[NoMoreExams]

so easy? are you sure?

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.

 

[trebmling]

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

 

[NoMoreExams]

Maybe it is, consult FTC

 

[trebmling]

what's FTS?

 

[NoMoreExams]

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

 

[trebmling]

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

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.

 

[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

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

 

[trebmling]

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

 

[NoMoreExams]

nope what?

 

[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.

 

[trebmling]

nope what?

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

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

 

[trebmling]

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