Use areas to evaluate ∫2−1(1−2x)dx∫−12(1−2x)dx.

A.

33

B.

55

C.

22

D.

1010

E.

00

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.

Use areas to evaluate ∫2−1(1−2x)dx ∫−12(1−2x)dx. A. 3 B. 5 C. 2 D. 10 E. 0

Learn how to evaluate definite integrals of the function (1 - 2x) over the limits from -1 to 2. This step-by-step solution applies the Fundamental Theorem of Calculus to calculate the area under the curve. Suitable for Grades 11-12 students.

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