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

This procedure eventually leads to the formula $$ \int\limits_a^b {f(x)dx = \sum\limits_{n = 1}^\infty {A_n } = \left( {b - a} \right)} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^{2^n - 1} {\left( { - 1} \right)^{m + 1} } } 2^{ - n} f\left( {a + m(b - a)/2^n } \right) $$

References

1. http://arxiv.org/abs/math.CA/0011078.
2. Int. J. Math. Math. Sci. 31, 345-351, 2002.