Class activity, September 6 solutions

What will happen?

For each question, predict what will happen when the code is run. Then run the code and check whether your prediction was correct.

1. Since x is 10, g02(x) returns 11. But values defined inside a function don’t impact the global environment, so calling g02 doesn’t change the value of x. Therefore x + 1 still returns 11.

x <- 10

g02 <- function(x){
  x <- x + 1
  return(x)
}

g02(x)

[1] 11

x + 1

[1] 11

2. Since x is 10, g02(x) returns 11. This output is stored in x, so the variable x has been overwritten and is now 11. When we run x + 1, we therefore get 12.

x <- 10

g02 <- function(x){
  x <- x + 1
  return(x)
}

x <- g02(x)
x + 1

[1] 12

3. When functions are nested inside each other, the inner function evaluates first. So, first R calculates g02(20), which is 19. This output is then used directly as the input for a second call to g02, and g02(19) returns 18.

g02 <- function(y){
  y <- y - 1
  return(y)
}

g02(g02(20))

[1] 18

Practice with anonymous functions

4. Use the integrate() and an anonymous function to find the area under the curve for the following functions:

• y = x^2 - x for x in \([0, 1]\)
• y = sin(x) + cos(x) for x in \([-\pi, \pi]\)
• y = exp(x)/x for x in \([10, 20]\)

integrate(function(x) {x^2 - x}, 0, 1)

-0.1666667 with absolute error < 1.9e-15

integrate(function(x) {sin(x) + cos(x)}, -pi, pi)

5.231803e-16 with absolute error < 6.3e-14

integrate(function(x) {exp(x)/x}, 10, 20)

25613160 with absolute error < 2.8e-07