Integration by Sampling the Integrand

An alternative approach is to choose randomly just the independent variable $x$. A series of random points $x_{i}$ produces a series of sample values $y_{i} = f(x_{i})$ of the integrand. If the points are chosen uniformly on the interval $[a,b]$ then the integral we want is just

$$I = <y>(b-a)$$ (7)

where

$$<y> = \sum_{i=1}^{N} y_{i}/N.$$ (8)

Now the error in the result is just the error in the estimate of the mean value of $y_{i}$:

$$\sigma_{<y>} = \sqrt{(<y^{2}> - <y>^{2})/N}.$$ (9)

so that

$$I = (b-a)<y> \pm (b-a)\sqrt{(<y^{2}> - <y>^{2})/N}.$$ (10)

If we consider the probability distribution $P(y)dy$ for getting a value in the range $[y,y+dy]$, then what we are trying to do is to determine the mean and standard deviation of this probability distribution. Notice that if the function values are definitely bounded by $y \in [y_{0},y_{1}]$, then the standard deviation of the population cannot exceed the allowed width of the distribution, so we have a rigorous upper bound on the error in the mean, and can plan our sample size accordingly: $\sigma_{<y>} \le \vert y_{1}-y_{0}\vert/\sqrt{N}$.