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The problem asks for the expression that calculates the area limited by the curve \( y = -2x + 3 \) from \( x = -2 \) to \( x = 1 \).

The area under a curve between two points is given by the integral of the function with respect to \( x \), i.e.,

\[
\text{Area} = \int_{a}^{b} f(x) \, dx
\]

Here, the curve is defined by \( y = -2x + 3 \). Therefore, the area calculation would be the integral of \( -2x + 3 \) from \( x = -2 \) to \( x = 1 \).

Thus, the correct expression is:

\[
\int_{-2}^{1} (-2x + 3) \, dx = \int_{-2}^{1} (-2x) \, dx + \int_{-2}^{1} 3 \, dx
\]

Looking at the options:

- Option **a** matches this form, as it breaks down the integral into two parts: \( \int_{-2}^{1} (-2x) \, dx \) and \( \int_{-2}^{1} 3 \, dx \).
- This corresponds to the correct solution.

Therefore, **the correct answer is option a**.

Would you like more details or have any further questions?

Here are 5 related questions:
1. How do you calculate the area under a curve using integrals?
2. What is the geometric interpretation of the integral of a linear function?
3. How do definite integrals account for areas below the x-axis?
4. What is the process of splitting integrals for a piecewise function?
5. Can the integral of a constant function over an interval be visualized as a rectangle?

**Tip:** When calculating the area between a curve and the x-axis, remember to check if the curve dips below the x-axis, as it may require adjusting for negative areas.

¿Cuál es la expresión que permite calcular el área limitada por la curva y = −2x + 3 desde x = −2 hasta x = 1?

Learn how to calculate the area under the curve y = −2x + 3 from x = −2 to x = 1 using definite integrals. This problem explores linear functions and the application of the Fundamental Theorem of Calculus to calculate the area between the curve and the x-axis.

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