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

• trebmling
In summary, the Fundamental Theorem of Calculus states that two different functions, such as G(x) and F(x), can be differentiated to get a new function, F'(x). This new function is the derivative of the original function. However, this differentiation may not be as easy as it seems.
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!

Differentiate both sides?

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

NoMoreExams said:
Differentiate both sides?

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

so easy? are you sure?

trebmling said:
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.

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

Maybe it is, consult FTC

what's FTS?

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

NoMoreExams said:
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.

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

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

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

Last edited:
nope what?

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.

NoMoreExams said:
nope what?

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}

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

## 1. What is an integral equation?

An integral equation is an equation that involves an unknown function within an integral. It represents the relationship between the unknown function and its derivative.

## 2. How do you solve an integral equation?

To solve an integral equation, you need to use techniques from integral calculus, such as integration by parts, substitution, or partial fractions. You can also use numerical methods, such as Simpson's rule or the trapezoidal rule.

## 3. What is the difference between G(x) and F(x) in an integral equation?

G(x) and F(x) are both functions that appear in the integral equation. G(x) represents the unknown function, which is to be solved for, while F(x) represents a known function. Together, they represent the relationship between the unknown function and its derivative.

## 4. Can an integral equation have multiple solutions?

Yes, an integral equation can have multiple solutions. This is because there can be multiple functions that satisfy the given relationship between the unknown function and its derivative. The uniqueness of the solution depends on the specific conditions and constraints of the integral equation.

## 5. What are the applications of solving integral equations?

Integral equations have numerous applications in physics, engineering, and other fields of science. They are commonly used to model physical systems and phenomena, such as heat transfer, diffusion, and fluid flow. They also play a crucial role in solving differential equations and boundary value problems.

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