PlanetPhysics/Fundamental Theorem of Integral Calculus

Consider the sequence of numbers ${\displaystyle \{f_{0},f_{1},...f_{N}\}}$ and define the difference ${\displaystyle \Delta f_{j}=f_{j}-f_{j-1}}$. Now sum the differences and not that all but the first and last terms cancel:

${\displaystyle \sum \Delta f_{j}=(f_{1}-f_{0})+(f_{2}-f_{1})+(f_{3}-f_{2})+...+(f_{N-2}-f_{N-1})+(f_{N}-f_{N-1})=f_{N}-f_{0}}$

In other words ${\displaystyle \int _{a}^{b}df=f(b)-f(a)}$. It seems obvious that,

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

Changing variables:

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

or as an indefinite integral:

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

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

${\displaystyle g(x)=\int _{a}^{x}f(s)ds=\sum _{0}^{N}f_{j}\Delta x=\left(f_{0}+f_{1}+...+f_{N}\right)\Delta x}$ where ${\displaystyle f_{N}=f(x)}$.

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

${\displaystyle g(x+\Delta x)=\int _{a}^{x+\Delta x}f(s)ds}$

${\displaystyle =\sum _{0}^{N+1}f_{j}\Delta x=\underbrace {\sum _{0}^{N}f_{j}\Delta x} _{g(x)}+f_{N+1}\Delta x}$.

${\displaystyle g(x+\Delta x)=g(x)+f(x+\Delta x)\Delta x}$.

Rearrange this to obtain:

${\displaystyle f(x+\Delta x)\Delta x=g(x+\Delta x)-g(x)}$

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