derivation of a definite integral formula using the method of exhaustion

The area under an arbitrary function $f(x)$ that is piecewise continuous on $[a,b]$ can be ”exhausted” with triangles. The first triangle has vertices at $(a,0)$ and $(b,0)$, and intersects $f(x)$ at

$$x=a+\frac{b-a}{2},$$

yielding the estimate

$$A_1=\frac{1}{2}(b-a)f(a+\frac{b-a}{2})$$

The second approximation involves two triangles, each sharing two vertices with the original triangle, and intersecting $f(x)$ at

$$x=a+\frac{b-a}{4}$$

and

$$x=a+\frac{3(b-a)}{4},$$

adding the area:

$$A_2=\frac{1}{4}(b-a)\{f(a+\frac{b-a}{4})-f(a+\frac{b-a}{2})+f(a+\frac{3(b-a)}{4})\}$$

A third such approximation involves four more triangles, adding the area

$$\begin{array}{c}\hfill A_3=\frac{1}{8}(b-a)\{f(a+\frac{b-a}{8})-f(a+\frac{b-a}{4})\hfill \\ \hfill +f(a+\frac{3(b-a)}{8})-f(a+\frac{b-a}{2})+f(a+\frac{5(b-a)}{8})\hfill \\ \hfill -f(a+\frac{3(b-a)}{4})+f(a+\frac{7(b-a)}{8})\}.\hfill \end{array}$$

This procedure eventually leads to the formula

$$\underset{a}{\overset{b}{\int }}f(x)𝑑x=\sum _{n=1}^{\mathrm{\infty }}{A}_n=\left(b-a\right)\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f\left(a+m(b-a)/2^n\right)$$

References

1. http://arxiv.org/abs/math.CA/0011078http://arxiv.org/abs/math.CA/0011078.

2. Int. J. Math. Math. Sci. 31, 345-351, 2002.