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 $\int _{a}^{b}df=f(b)-f(a)$. 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:

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

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

As shown in red, the change in area (∫fdx) of a function is closely related to the value of the function, f(x) at the point where x changes to x+δx

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

Rearrange this to obtain:

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

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