1. We need to count all 4-digit integers in the interval $[2022,4482]$ using only the digits $\{0,2,3,4,6,7\}.$

2. Since the number must lie between $2022$ and $4482$, let the 4-digit number be $abcd$. We count valid numbers casewise by the thousands digit.

Case 1: Thousands digit = $2$

Then the number is of the form $2bcd$. To be in $[2022,4482]$, it must satisfy $2bcd \ge 2022.$

All digits must come from $\{0,2,3,4,6,7\}$.

Total numbers starting with $2$:

• $b$: $6$ choices
• $c$: $6$ choices
• $d$: $6$ choices

So total is $6^3=216.$

But some are less than $2022$. Since the first digit is already $2$, compare with $2022$:

• If $b=0$, then compare the last two digits with $22$.
  • If $c=0$, then any $d$ gives numbers $200d < 2022$. This contributes $6$ invalid numbers.
  • If $c=2$, then we need $d\ge 2$ to be valid. So only $d=0$ is invalid. This contributes $1$ invalid number.
  • If $c\in\{3,4,6,7\}$, then number is already $>2022$.

Thus invalid numbers below $2022$ are $6+1=7.$

Hence valid numbers in this case: $216-7=209.$

Case 2: Thousands digit = $3$

Any number of the form $3bcd$ is automatically between $2022$ and $4482$.

Choices for each of $b,c,d$ are $6$ each, so count is $6^3=216.$

Case 3: Thousands digit = $4$

Now the number is of the form $4bcd$, and must satisfy $4bcd \le 4482.$

We count valid possibilities by comparing with $4482$.

Subcase 3.1: $b<4$

Allowed digits less than $4$ from the set are $0,2,3.$ So $b$ has $3$ choices. Then $c,d$ can be anything from the set: $6$ choices each.

Count: $3\cdot 6\cdot 6=108.$

Subcase 3.2: $b=4$

Now compare $44cd \le 4482$.

So we need $cd \le 82.$ Both digits must be from $\{0,2,3,4,6,7\}$.

• If $c<8$, then since available digits for $c$ are all from the set, all possible $c\in\{0,2,3,4,6,7\}$ are less than $8$. For each such $c$, any $d$ is allowed. That gives $6\cdot 6=36.$

But when $c=8$ would need $d\le 2$, impossible since $8$ is not available. So no extra restriction arises.

Thus count here is $36$.

Hence total for thousands digit $4$ is $108+36=144.$

4. Therefore total count is $209+216+144=569.$

So the required number of integers is $\boxed{569}.$

Comparison with stored answer

Stored correct answer = $569$.

Our derived answer is also $569$, so they agree.