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When I approximate the derivative with forward, backward and central difference how can I use definite integral to get back the original function. Assume I have a function.

$$x(t)=t^2$$

provided $ x(0) = 0 $

Then I approximate the derivative with backward method (with $\Delta$t = 1), I get $$x '(t) = x(t) - x(t-1) = t^2 - (t-1) ^2 = t^2-(t^2-2t+1) = 2t-1$$

Then I want to take definite integral: $$\int_0^t{x ' (t)}$$ in this case I get back $$ x(t) =t^2-t$$

how can I get back the original function? is there other way to do integral? or do I have to make some adjustment?

please give me detail and step by step easy to understand because I'm newbie. My knowledge is only calculus 2 level (Taylor series level).

please also give answer to all three method Forward, Backward, and Central Difference of how to get back the original function by using definite integral.

I'm studying Linear algebra, the book has mentioned about the invertibility of difference matrix and connect it with the idea of calculus.

The book says:

$$A= \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$

$$x= \begin{bmatrix} 1 \\ 4 \\ 9 \end{bmatrix}$$

$$b= \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}$$

$$S= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$$

Then

$$Ax = b$$

$$x= Sb$$

The book say the matrix A and S is like differentiation and integration of a function. and x is like a function x(t).

so if be do Ax we get b and if we do Sb beget back x. It is like

$$\frac{dx}{dt} = b$$ and $$x(t) = \int_0^b{b}$$

but if $x(t) = x^2$ we need to approximate the derivative with backward method to get x '(t) = 2t-1 = vector b

after we get vector b by doing backward approximate. How can I get back to vector x or in other word x(t) by some type of integration?

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  • $\begingroup$ Note that $x'(t)$ is not the same as $x(t)-x(t-1)$. The latter is simply an approximation. $\endgroup$ – preferred_anon Aug 11 '16 at 8:57
  • $\begingroup$ It's our destiny: integration is defined only with constant, you can't avoid it $\endgroup$ – Anna Abasheva Aug 12 '16 at 19:10

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