To determine the number of points of discontinuity for the function [tex]\(\phi(x) = \int_0^x [\sqrt{t}] \, dt\)[/tex], where [tex]\([r]\)[/tex] denotes the greatest integer less than or equal to [tex]\(r\)[/tex], we need to understand the behavior of the integrand [tex]\([\sqrt{t}]\)[/tex].
1. Understanding the Function [tex]\([\sqrt{t}]\)[/tex]:
- The function [tex]\([\sqrt{t}]\)[/tex] represents the greatest integer function applied to [tex]\(\sqrt{t}\)[/tex]. This function is essentially a step function because it takes integer values and jumps at certain points.
2. Discontinuities of [tex]\([\sqrt{t}]\)[/tex]:
- The function [tex]\([\sqrt{t}]\)[/tex] changes its value discontinuously at points where [tex]\(\sqrt{t}\)[/tex] is an integer.
- Specifically, [tex]\([\sqrt{t}]\)[/tex] jumps at [tex]\(t = n^2\)[/tex] where [tex]\(n\)[/tex] is a positive integer. At these points, [tex]\(\sqrt{t}\)[/tex] equals [tex]\(n\)[/tex], causing [tex]\([\sqrt{t}]\)[/tex] to increase by 1.
3. Range of [tex]\(t\)[/tex]:
- We are considering the interval [tex]\(0 \leq x \leq 2023\)[/tex]. Therefore, we need to determine for which values of [tex]\(t\)[/tex] in [tex]\(0 \leq t \leq 2023\)[/tex] the function [tex]\([\sqrt{t}]\)[/tex] changes its value.
- We need the squares of integers [tex]\(n\)[/tex] such that [tex]\(n^2 \leq 2023\)[/tex].
4. Maximum Value of [tex]\(n\)[/tex]:
- To find the maximum integer [tex]\(n\)[/tex] such that [tex]\(n^2 \leq 2023\)[/tex], we solve for [tex]\(n\)[/tex]:
[tex]\[
n \leq \sqrt{2023}
\][/tex]
- Calculating, we find:
[tex]\[
\sqrt{2023} \approx 44.94
\][/tex]
- Thus, the largest integer [tex]\(n\)[/tex] is 44 because [tex]\(44^2 = 1936 \leq 2023\)[/tex] and [tex]\(45^2 = 2025 > 2023\)[/tex].
5. Number of Discontinuities:
- The points [tex]\(t = n^2\)[/tex] for [tex]\(n = 1, 2, 3, \ldots, 44\)[/tex] are the points where [tex]\([\sqrt{t}]\)[/tex] jumps and hence [tex]\(\phi(x)\)[/tex] has discontinuities.
- Therefore, the number of points of discontinuity is the number of integers from 1 to 44, which is exactly 44.
So, the number of points of discontinuity is:
[tex]\[
\boxed{44}
\][/tex]
