The problem provides a table of mass intervals and corresponding frequencies. We need to find the smallest integer value that the frequency density axis needs to reach in order to plot all of the data in a histogram. Frequency density is calculated as frequency divided by class width.

1. Problem Description

The problem provides a table of mass intervals and corresponding frequencies. We need to find the smallest integer value that the frequency density axis needs to reach in order to plot all of the data in a histogram. Frequency density is calculated as frequency divided by class width.

2. Solution Steps

First, we need to calculate the class width for each interval. Then, we calculate the frequency density for each interval. Finally, we find the maximum frequency density and round it up to the nearest integer.

Class Width:

$0 < x \le 14$: Class width = $14 - 0 = 14$

$14 < x \le 18$: Class width = $18 - 14 = 4$

$18 < x \le 20$: Class width = $20 - 18 = 2$

$20 < x \le 25$: Class width = $25 - 20 = 5$

$25 < x \le 40$: Class width = $40 - 25 = 15$

Frequency Density:

Frequency Density = $\frac{Frequency}{Class Width}$

$0 < x \le 14$: Frequency Density = $\frac{21}{14} = 1.5$

$14 < x \le 18$: Frequency Density = $\frac{28}{4} = 7$

$18 < x \le 20$: Frequency Density = $\frac{17}{2} = 8.5$

$20 < x \le 25$: Frequency Density = $\frac{22}{5} = 4.4$

$25 < x \le 40$: Frequency Density = $\frac{33}{15} = 2.2$

The maximum frequency density is $8.5$.

The smallest integer value that the frequency density axis needs to reach is the smallest integer greater than or equal to the maximum frequency density, so we round $8.5$ up to the nearest integer, which is $9$.

3. Final Answer

9