We need to find the sum of all even integers from 20 to 40, inclusive. Let's denote this sum by $x$.
Step 1: Identify the Sequence
The even integers from 20 to 40 form an arithmetic sequence:

$$
\(20,22,24, \ldots, 40\)
$$


Step 2: Determine the Number of Terms
An arithmetic sequence has the general form:

$$
\(a_n=a_1+(n-1) d\)
$$

where:
- $\(a_n\)$ is the $n$-th term,
- $\(a_1\)$ is the first term,
- $\(d\)$ is the common difference,
- $\(n\)$ is the number of terms.

Given:
- First term $\(\left(a_1\right)=20\)$,
- Last term $\(\left(a_n\right)=40\)$,
- Common difference $\((d)=2\)$.

To find the number of terms ( $n$ ):

$$
\(\begin{gathered}
40=20+(n-1) \times 2 \\
40-20=(n-1) \times 2 \\
20=(n-1) \times 2 \\
n-1=10 \Longrightarrow n=11
\end{gathered}\)
$$

Step 3: Calculate the Sum of the Sequence
The sum $S$ of an arithmetic sequence is given by:

$$
\(S=\frac{n}{2} \times\left(a_1+a_n\right)\)
$$


Substituting the known values:

$$
\(x=\frac{11}{2} \times(20+40)=\frac{11}{2} \times 60=11 \times 30=330\)
$$


Verification

Alternatively, we can verify by pairing terms:

$$
\((20+40)+(22+38)+(24+36)+\ldots\)
$$


Each pair sums to 60, and there are $\(\frac{11}{2}=5.5\)$ such pairs. Multiplying:

$$
\(5.5 \times 60=330\)
$$


Final Answer

The sum of the even integers from 20 to 40, inclusive, is:
330