$\theta=\int_{0}^{2}3e^{-3x}dx$
Using Monte-Carlo method estimate the confidence intervals for the integral above. Use a distribution different from the uniform distribution to minimize the confidence intervals.
I tried using variable substitute so $\theta=\int_{0}^{1}6e^{-6t}dt$ and using the beta distribution(1,6) the plots seem to agree but the estimate for $\theta$ seems to be way to big, here is the R output:
> integral2<-function(n) {
+ x<-runif(n)
+ fy<-dbeta(x,1,6)
+ g_y<-6*exp(1)^(-6*x)
+ hy<-(g_y/fy)
+ sdg_y<-sd(g_y)
+ theta<-mean(hy)
+ sdtheta<-sqrt(sum((hy-theta)^2)/(n-1))
+ print(max(hy))
+ print(theta)
+ print(sqrt(sum((hy-theta)^2)/(n-1)))
+
+ return(c(theta-sdtheta*qnorm(0.975)/sqrt(n),theta+sdtheta*qnorm(0.975)/sqrt(n)))
+ }
> integral2(10000)
[1] 1.484229e+17 #max(hy)
[1] 2.123842e+13 #theta
[1] 1.608726e+15 #sdtheta
[1] -1.029203e+13 5.276887e+13 #confidence interval for theta