Monte-Carlo for Integrals

Let $X$ be a Random Variable with Uniform Distribution on $(0,1)$ and $f$ a continous function on $[0,1]$ , then we can calculate the intergral from $0$ to $1$ of that function using the Arithmetic Mean as a Standard Estimator: $\mathbb{E}(f(X))={\int }_0^{1}f(x)dx$

This approach is quite effective for high-dimensional integrals.

x = runif(100000)
mean(sin(x)) - integrate(sin, 0, 1)[1]$value
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