How to use standard Monte Carlo to estimate \(\theta = \int_a^b g(x) dx\) when \((a,b) \neq (0,1)\)?
When \((a,b) = (0,1)\), we generate \(U \sim \textrm{Unif}(0,1)\) as the underlying random variable in standard Monte Carlo since the support of \(U\) is same as interval of integration. The idea is essentially \[ \int_0^1 g(x) dx = \int_0^1 g(u) f(u) du = \mathbb{E}\{g(U)\}, \] since the pdf of \(U\) is \(f(u)=1\).
Similarly, for \(\theta = \int_a^b g(x) dx\), we generate \(V \sim U(a,b)\). Then \[ \int_a^b g(x) dx = (b-a)\int_a^b g(v) \frac{1}{b-a} dv = (b-a)\mathbb{E}\{g(V)\}, \] since the pdf of \(V\) is \(f(v)=\frac{1}{b-a}\). This means we need to multiply \((b-a)\) to the original sample average from simulation.
The following is an illustration with estimation of \(\theta = \int_{1.7}^{7.1} e^x dx\):
n = 100000
a = 1.7
b = 7.1
U = runif(n, a, b)
thetaH = mean(exp(U))
thetaH #original sample average## [1] 220.9045(b-a)*thetaH #adjusted sample average## [1] 1192.885exp(b)-exp(a) #theoretical value## [1] 1206.493