PlanetPhysics/Fundamental Theorem of Integral Calculus

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

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

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

$\underset{a}{\overset{b}{\int }}\frac{df}{dx}dx=\int df=f(b)-f(a)$

Changing variables:

$\underset{a}{\overset{x}{\int }}\frac{df}{ds}=\int df=f(x)-f(a)$

or as an indefinite integral:

$\int f^{\prime }(x)dx=f(x)+C$

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,

$g(x)=\underset{a}{\overset{x}{\int }}f(s)ds=\sum _0^N{f}_j\mathrm{\Delta }x=(f_0+f_1+...+f_N)\mathrm{\Delta }x$ where $f_N=f(x)$.

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

$g(x+\mathrm{\Delta }x)=\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(s)ds$

$=\sum _0^{N+1}{f}_j\mathrm{\Delta }x=\underset{g(x)}{\underbrace{\sum _0^N{f}_j\mathrm{\Delta }x}}+f_{N+1}\mathrm{\Delta }x$.

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 Template:Nowrap

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

Rearrange this to obtain:

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

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