The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ______.

The permitted digit set is $$\{0,\,2,\,3,\,4,\,6,\,7\}$$ (six digits).
Every required integer is 4-digit and must lie in the closed interval $$[2022,\,4482]$$.

Denote the thousands, hundreds, tens and units digits by $$T,H,E,U$$ respectively.
Because $$2022\le T\,H\,E\,U\le 4482$$, the thousands digit $$T$$ can take only the values $$2,3,4$$ (digits $$0,6,7$$ are either too small or would exceed $$4482$$).

The smallest admissible number is $$2022$$, so the last three-digit part $$H\,E\,U$$ must satisfy $$H\,E\,U\ge 022$$.

Total 3-digit strings possible with the allowed digits: $$6^3 = 216$$.

We must exclude those strings for which $$H\,E\,U\lt 022$$, i.e. $$000$$ to $$021$$. Among these first 22 numbers, only the following possess digits solely from the permitted set: $$000,002,003,004,006,007,020$$  —  7 strings in all.

Hence valid numbers when $$T=2$$: $$216-7 = 209$$.

All numbers $$3000$$-$$3999$$ automatically lie inside the interval. Each of $$H,E,U$$ can again be any of the 6 allowed digits, giving $$6^3 = 216$$ valid numbers.

Now the highest permissible integer is $$4482$$, so the last three-digit part must satisfy $$H\,E\,U\le 482$$.

Total possible 3-digit strings: $$6^3 = 216$$.
We count directly how many of these are $$\le 482$$.

The hundreds digit $$H$$ can be $$0,2,3,4$$ (digits $$6,7$$ would already yield $$\ge 600$$).

For $$H=0,2,3$$ every choice of $$E,U$$ works (because the number is $$\le 399$$). That yields $$3\times 6\times 6 = 108$$ numbers.

For $$H=4$$ we need $$4E U \le 482 \Longrightarrow 10E+U\le 82$$.
For each allowed tens digit $$E=0,2,3,4,6,7$$ the inequality $$U\le 82-10E$$ is satisfied by all six allowed units digits (each is at most $$7$$). Thus $$6\times 6 = 36$$ additional numbers.

Total for $$T=4$$: $$108+36 = 144$$.

Adding the three cases:

$$209 + 216 + 144 = 569$$

Therefore, the number of 4-digit integers required is 569.