# one-dimensional integrals

December 25, 2010
By

(This article was first published on YGC » R, and kindly contributed to R-bloggers)

The foundamental idea of numerical integration is to estimate the area of the region in the xy-plane bounded by the graph of function f(x). The integral was esimated by divide x to small intervals, then add all the small approximations to give a total approximation.

Numerical integration can be done by trapezoidal rule, simpson’s rule and quadrature rules. R has a built-in function integrate, which performs adaptive quadrature.

Trapezoidal rule works by approximating the region under the graph f(x) as a trapezoid and calculating its area.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 ``` ```trapezoid <- function(fun, a, b, n=100) { # numerical integral of fun from a to b # using the trapezoid rule with n subdivisions # assume a < b and n is a positive integer h <- (b-a)/n x <- seq(a, b, by=h) y <- fun(x) s <- h * (y[1]/2 + sum(y[2:n]) + y[n+1]/2) return(s) }```

Simpson’s rule subdivides the interval [a,b] into n subintervals, where n is even, then on each consecutive pairs of subintervals, it approximates the behaviour of f(x) by a parabola (polynomial of degree 2) rather than by the straight lines used in the trapezoidal rule.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 ``` ```simpson <- function(fun, a, b, n=100) { # numerical integral using Simpson's rule # assume a < b and n is an even positive integer h <- (b-a)/n x <- seq(a, b, by=h) if (n == 2) { s <- fun(x[1]) + 4*fun(x[2]) +fun(x[3]) } else { s <- fun(x[1]) + fun(x[n+1]) + 2*sum(fun(x[seq(2,n,by=2)])) + 4 *sum(fun(x[seq(3,n-1, by=2)])) } s <- s*h/3 return(s) }```

To calculate an integrate over infinite interval, one way is to transform it into an integral over a finite interval as introduce in wiki.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ``` ```simpson_v2 <- function(fun, a, b, n=100) { # numerical integral using Simpson's rule # assume a < b and n is an even positive integer if (a == -Inf & b == Inf) { f <- function(t) (fun((1-t)/t) + fun((t-1)/t))/t^2 s <- simpson_v2(f, 0, 1, n) } else if (a == -Inf & b != Inf) { f <- function(t) fun(b-(1-t)/t)/t^2 s <- simpson_v2(f, 0, 1, n) } else if (a != -Inf & b == Inf) { f <- function(t) fun(a+(1-t)/t)/t^2 s <- simpson_v2(f, 0, 1, n) } else { h <- (b-a)/n x <- seq(a, b, by=h) y <- fun(x) y[is.nan(y)]=0 s <- y[1] + y[n+1] + 2*sum(y[seq(2,n,by=2)]) + 4 *sum(y[seq(3,n-1, by=2)]) s <- s*h/3 } return(s) }```
```> phi <- function(x) exp(-x^2/2)/sqrt(2*pi)  ##normal distribution
> simpson_v2(phi, 0, Inf, n=100)
[1] 0.4986569
> simpson_v2(phi, -Inf,3, n=100)
[1] 0.998635
> pnorm(3)
[1] 0.9986501
```

Simpson’s rule is more accuracy than trapezoidal rule. To compare the accuracy between simpson’s rule and trapezoidal rule, I estimated $int_{0.01}^{1} 1/x dx$ = -log(0.01) for a sequence of increasing values of n.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ```f <- function(x) 1/x #integrate(f, 0.01, 1) == -log(0.01) S.trapezoid <- function(n) trapezoid(f, 0.01, 1, n) S.simpson <- function(n) simpson(f, 0.01, 1, n)   n <- seq(10, 1000, by = 10) St <- sapply(n, S.trapezoid) Ss <- sapply(n, S.simpson)   opar <- par(mfrow = c(2, 2)) plot(n,St + log(0.01), type="l", xlab="n", ylab="error", main="Trapezoidal rule") plot(n,Ss + log(0.01), type="l", xlab="n", ylab="error",main="Simpson's rule") plot(log(n), log(St+log(0.01)),type="l", xlab="log(n)", ylab="log(error)",main="Trapezoidal rule") plot(log(n), log(Ss+log(0.01)),type="l", xlab="log(n)", ylab="log(error)",main="Simpson's rule")```

The plot showed that log(error) against log(n) appears to have a slope of -1.90 and -3.28 for trapezoidal rule and simpson’s rule respectively.

Another way to compare their accuracy is to calculate how large of the partition size n for reaching a specific tolerance.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 ``` ```yIntegrate <- function(fun, a, b, tol=1e-8, method= "simpson", verbose=TRUE) { # numerical integral of fun from a to b, assume a < b # with tolerance tol n <- 4 h <- (b-a)/4 x <- seq(a, b, by=h) y <- fun(x) yIntegrate_internal <- function(y, h, n, method) { if (method == "simpson") { s <- y[1] + y[n+1] + 4*sum(y[seq(2,n,by=2)]) + 2 *sum(y[seq(3,n-1, by=2)]) s <- s*h/3 } else if (method == "trapezoidal") { s <- h * (y[1]/2 + sum(y[2:n]) + y[n+1]/2) } else { } return(s) }   s <- yIntegrate_internal(y, h, n, method) s.diff <- tol + 1 # ensures to loop at once. while (s.diff > tol ) { s.old <- s n <- 2*n h <- h/2 y[seq(1, n+1, by=2)] <- y ##reuse old fun values y[seq(2,n, by=2)] <- sapply(seq(a+h, b-h, by=2*h), fun) s <- yIntegrate_internal(y, h, n, method) s.diff <- abs(s-s.old) } if (verbose) { cat("partition size", n, "n") } return(s) }```
```> fun <- function(x) exp(x) - x^2
> yIntegrate(fun, 3,5,tol=1e-8, method="simpson")
partition size 512
[1] 95.66096
> yIntegrate(fun, 3,5,tol=1e-8, method="trapezoidal")
partition size 131072
[1] 95.66096
```

As show above, simpson’s rule converge much faster than trapezoidal rule.

Here, I presented the adaptive simpson’s method.

?View Code RSPLUS
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ``` ```quadrature <- function(fun, a, b, tol=1e-8) { # numerical integration using adaptive quadrature   quadrature_internal <- function(S.old, fun, a, m, b, tol, level) { level.max <- 100 if (level > level.max) { cat ("recursion limit reached: singularity likelyn") return (NULL) } S.left <- simpson(fun, a, m, n=2) S.right <- simpson(fun, m, b, n=2) S.new <- S.left + S.right if (abs(S.new-S.old) > tol) { S.left <- quadrature_internal(S.left, fun, a, (a+m)/2, m, tol/2, level+1) S.right <- quadrature_internal(S.right, fun, m, (m+b)/2, b, tol/2, level+1) S.new <- S.left + S.right } return(S.new) }   level = 1 S.old <- (b-a) * (fun(a) + fun(b))/2 S.new <- quadrature_internal(S.old, fun, a, (a+b)/2, b, tol, level+1) return(S.new) }```

Quadrature rule is effective when f(x) is steep.

```> fun <- function(x) return(1.5 * sqrt(x))
> system.time(yIntegrate(fun, 0,1, tol=1e-9, method="simpson"))
partition size 524288
user  system elapsed
1.58    0.00    1.58
user  system elapsed
0.25    0.00    0.25
```

Reference:
1. Robinson A., O. Jones, and R. Maillardet. 2009. Introduction to Scientific Programming and Simulation Using R. Chapman and Hall.
2. http://en.wikipedia.org/wiki/Numerical_integration
3. http://en.wikipedia.org/wiki/Trapezoidal_rule
4. http://en.wikipedia.org/wiki/Simpson%27s_rule

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