# 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

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

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

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

 $\begin{array}[]{c}A_{3}{\rm{}}=\frac{{{\rm{}}1}}{8}(b-a)\{f(a+\frac{b-a}{8})-f% (a+\frac{b-a}{4})\\ +f(a+\frac{3(b-a)}{8})-f(a+\frac{b-a}{2})+f(a+\frac{5(b-a)}{8})\\ -f(a+\frac{3(b-a)}{4})+f(a+\frac{7(b-a)}{8})\}.\\ \end{array}$
 $\int\limits_{a}^{b}{f(x)dx=\sum\limits_{n=1}^{\infty}{A_{n}}=\left({b-a}\right% )}\sum\limits_{n=1}^{\infty}{\sum\limits_{m=1}^{2^{n}-1}{\left({-1}\right)^{m+% 1}}}2^{-n}f\left({a+m(b-a)/2^{n}}\right)$

References

1. 1.

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

2. 2.

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

Title derivation of a definite integral formula using the method of exhaustion DerivationOfADefiniteIntegralFormulaUsingTheMethodOfExhaustion 2013-03-22 14:56:35 2013-03-22 14:56:35 ruffa (7723) ruffa (7723) 22 ruffa (7723) Derivation msc 78A45 msc 30B99 msc 26B15