PlanetPhysics/Fundamental Theorem of Integral Calculus

Consider the sequence of numbers $$\{f_0,f_1,...f_N\}$$ and define the difference $$\Delta f_j = f_j-f_{j-1}$$. Now sum the differences and not that all but the first and last terms cancel:

$$\sum \Delta f_j = (f_1-f_0) + (f_2-f_1) + (f_3-f_2) + ... + (f_{N-2} - f_{N-1}) + (f_N - f_{N-1}) = f_N-f_0$$

In other words $$\int_a^bdf = f(b)-f(a)$$. It seems obvious that,

$$\int_a^b\frac{df}{dx}dx= \int df = f(b)-f(a)$$

Changing variables:

$$\int_a^x\frac{df}{ds}= \int df = f(x)-f(a)$$

or as an indefinite integral:

$$\int f'(x)dx = f(x) + C$$

Converse
In other words, the integral of the derivative of a function is the original function. But what of the derivative of the integral? Let,

$$g(x) = \int_a^x f(s)ds = \sum_0^N f_j \Delta x = \left(f_0+f_1+...+f_N \right)\Delta x$$ where $$f_N = f(x)$$.

Here, we assume that all the intervals $$\Delta x$$ in the Riemann sum are equal. To find $$g(x+\Delta x)$$ we need to add one extra term to the Riemann sum:

$$g(x+\Delta x) = \int_a^{x+\Delta x} f(s)ds $$
 * $$= \sum_0^{N+1} f_j \Delta x = \underbrace{\sum_0^{N} f_j \Delta x}_{g(x)} + f_{N+1}\Delta x$$.

$$g(x+\Delta x) = g(x) + f(x+\Delta x)\Delta x$$.

Rearrange this to obtain:

$$f(x+\Delta x)\Delta x = g(x+\Delta x) - g(x) $$

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