Let the four consecutive odd integers be represented as \( x, x+2, x+4, x+6 \). Here, \( x \) is the smallest of the four integers.
We can set up the equation for their sum:
\[ x + (x+2) + (x+4) + (x+6) = 32 \]
Combining like terms, we get:
\[ 4x + 12 = 32 \]
Next, we will solve for \( x \) by first subtracting 12 from both sides:
\[ 4x = 32 - 12 \]
\[ 4x = 20 \]
Now, dividing both sides by 4:
\[ x = \frac{20}{4} \]
\[ x = 5 \]
Thus, the least integer is \( x = 5 \).
To confirm, we can find the four consecutive odd integers:
- The integers are \( 5, 7, 9, 11 \).
Calculating their sum:
\[ 5 + 7 + 9 + 11 = 32 \]
Since the sum is indeed 32, the solution is verified. Therefore, the value of the least integer is
\[ \boxed{5} \]