Answer to Question #202753 in Real Analysis for Raj Kumar

Question

Let f [: − 3,3 ] → R be defined by f (x)= 5[x] + x³where [x] denotes the greatest integer ≤ x. Show that this function is integrable.

Answer 1

First, we show that the floor function [x] is integrable. To see this, we have to consider the value of the function for each sub-interval.

For -3≤x<-2, [x]=-3

For -2≤x<-1, [x]=-2

For -1≤x<0, [x]=-1

For 0≤x<1, [x]=0

For 1≤x<2, [x]=1

For 2≤x<3, [x] =2

For x=3, [x]=3.

From this, we have that the concept of area under a curve is reduced to area of a rectangle. And the sum of the area of these rectangles is the area of our floor function. Hence it is integrable. So also, 5[x] is also integrable.

Hence, f(x) is integrable because it is the addition of two integrable functions on [-3,3].