The sum of the first $x$ multiples of 11 can be written as:

$$
\(\begin{aligned}
& S(x)=11 \times 1+11 \times 2+11 \times 3+\cdots+11 \times x \\
& S(x)=11(1+2+3+\cdots+x)
\end{aligned}\)
$$


The sum of the first $x$ positive integers is given by the formula $\(\frac{x(x+1)}{2}$\).
So, $\(S(x)=11 \times \frac{x(x+1)}{2}\)$.
We need to find the value of $\mathrm{S}(40)$. Substitute $x=40$ into the formula:

$$
\(\begin{aligned}
& S(40)=11 \times \frac{40(40+1)}{2} \\
& S(40)=11 \times \frac{40 \times 41}{2} \\
& S(40)=11 \times(20 \times 41) \\
& S(40)=11 \times 820
\end{aligned}\)
$$


Now, perform the multiplication:

$$
\(\begin{aligned}
& 11 \times 820=11 \times(800+20) \\
& =11 \times 800+11 \times 20 \\
& =8800+220 \\
& =9020
\end{aligned}\)
$$


Let's check with the example given:

$$
\(S(20)=11+22+\ldots+220\)
$$


Using the formula:

$$
\(\begin{aligned}
& \mathrm{S}(20)=11 \times \frac{20(20+1)}{2} \\
& \mathrm{~S}(20)=11 \times \frac{20 \times 21}{2} \\
& \mathrm{~S}(20)=11 \times(10 \times 21) \\
& \mathrm{S}(20)=11 \times 210=2310 . \text { This matches the example. }
\end{aligned}\)
$$


Therefore, $\(S(40)=9020\)$.