Let f(x) = [x]^2 - 5[x-1] - 1 , where [t] denotes the greatest integer less than or equal to t . If ₀^ k |f(x)| d x = 18 , then the value of k is

Step-by-step solution

Simplify the function f(x) using the property [x-1] = [x] - 1 : f(x) = [x]^2 - 5([x] - 1) - 1 = [x]^2 - 5[x] + 4 Let n = [x] . Then f(x) = n^2 - 5n + 4 = (n-1)(n-4) . Since [x] is constant on intervals of the form [n, n+1) , f(x) is a piecewise constant step function. Evaluate |f(x)| for successive integer intervals: For x [0, 1), n = 0 |f(x)| = |(0-1)(0-4)| = 4 For x [1, 2), n = 1 |f(x)| = |(0)(-3)| = 0 For x [2, 3), n = 2 |f(x)| = |(1)(-2)| = 2 For x [3, 4), n = 3 |f(x)| = |(2)(-1)| = 2 For x [4, 5), n = 4 |f(x)|