To find the sum of the integers from 1 to 100 that are divisible by both 2 and 5, we need to find the sum of the arithmetic sequence of these numbers.

The first term of the sequence is 10 (the smallest number divisible by both 2 and 5), and the last term is 100.

The common difference between consecutive terms is 10 (since every number divisible by 2 and 5 is 10 greater than the previous one).

Using the formula for the sum of an arithmetic sequence, we have:

[S = \frac{n}{2}(a_1 + a_n)] where (S) is the sum, (n) is the number of terms, (a_1) is the first term, and (a_n) is the last term.

In this case, (n = \frac{a_n - a_1}{d} + 1 = \frac{100 - 10}{10} + 1 = 10 + 1 = 11).

Plugging in the values, we get:

[S = \frac{11}{2}(10 + 100) = \frac{11}{2}(110) = 11 \cdot 55 = \boxed{605}]

Therefore, the sum of the integers from 1 to 100 that are divisible by both 2 and 5 is 605.