We are tasked with evaluating the expression

$\int_{2}^{-1} (1 - 2x) \, dx \int_{-1}^{2} (1 - 2x) \, dx$

This consists of two integrals that need to be evaluated individually, and then we find the product of their results. Let’s compute each integral step-by-step.

Step 1: Evaluate $\int_{2}^{-1} (1 - 2x) \, dx$

We will first compute the integral $\int (1 - 2x) \, dx$:

$\int (1 - 2x) \, dx = x - x^2 + C$

Now, we apply the limits from 2 to -1:

[ \left[ x - x^2 \right]_{2}^{-1} = \left[ (-1) - (-1)^2 \right] - \left[ 2 - 2^2 \right] ] $= (-1 - 1) - (2 - 4) = (-2) - (-2) = -2 + 2 = 0$

So, the first integral evaluates to $0$.

Step 2: Evaluate $\int_{-1}^{2} (1 - 2x) \, dx$

We now compute the same integral, but with limits from -1 to 2:

[ \left[ x - x^2 \right]_{-1}^{2} = \left[ 2 - 2^2 \right] - \left[ (-1) - (-1)^2 \right] ] $= (2 - 4) - (-1 - 1) = (-2) - (-2) = -2 + 2 = 0$

So, the second integral also evaluates to $0$.

Step 3: Product of the two integrals

Since both integrals result in $0$, their product is:

$0 \times 0 = 0$

Thus, the value of the entire expression is $\boxed{0}$, which corresponds to option E.

---

Would you like further details on the method used?

Here are some additional questions to consider:

1. What is the geometric interpretation of the integral in this problem?
2. How would the result change if the function were different?
3. Can you describe how symmetry plays a role in integrals like these?
4. What happens if we reverse the limits of integration?
5. How do integrals of polynomial functions behave when evaluated over symmetric intervals?

Tip: When the result of an integral over symmetric limits is zero, it often suggests that the area above and below the x-axis cancels out.