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The problem asks for the area enclosed by the curve \( y = -x^2 - 3x + 4 \), the x-axis, and the lines \( x = -2 \) and the y-axis. To solve this, we need to calculate the area under the curve from \( x = -2 \) to \( x = 0 \).

### Step-by-Step Solution:

1. **Set up the integral**:
   The curve equation is \( y = -x^2 - 3x + 4 \). The area can be calculated by integrating this function between \( x = -2 \) and \( x = 0 \):

   \[
   A = \int_{-2}^{0} (-x^2 - 3x + 4) \, dx
   \]

2. **Compute the integral**:
   Let's integrate term by term:

   \[
   \int (-x^2) \, dx = -\frac{x^3}{3}
   \]
   \[
   \int (-3x) \, dx = -\frac{3x^2}{2}
   \]
   \[
   \int 4 \, dx = 4x
   \]

   Therefore, the integral becomes:

   \[
   A = \left[ -\frac{x^3}{3} - \frac{3x^2}{2} + 4x \right]_{-2}^{0}
   \]

3. **Evaluate at the limits**:
   Now substitute \( x = 0 \) and \( x = -2 \) into the expression:

   - At \( x = 0 \):

     \[
     -\frac{(0)^3}{3} - \frac{3(0)^2}{2} + 4(0) = 0
     \]

   - At \( x = -2 \):

     \[
     -\frac{(-2)^3}{3} - \frac{3(-2)^2}{2} + 4(-2) = -\frac{-8}{3} - \frac{12}{2} - 8
     \]
     Simplifying each term:
     \[
     \frac{8}{3} - 6 - 8 = \frac{8}{3} - 14
     \]

     Converting \( -14 \) to have a denominator of 3:
     \[
     \frac{8}{3} - \frac{42}{3} = \frac{-34}{3}
     \]

4. **Calculate the final area**:
   The total area is the absolute value of the result:

   \[
   A = \left| \frac{-34}{3} \right| = \frac{34}{3}
   \]

Thus, the area enclosed by the curve and the lines is \( \frac{34}{3} \), which corresponds to option ④.

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Would you like more details or have any questions?

### Relative questions:
1. What is the geometric meaning of integration when calculating areas?
2. How would you approach this problem if the curve crossed the x-axis?
3. Can this technique be used to find the area between two curves?
4. How do limits of integration affect the area calculation?
5. What would change if the curve's equation was different?

### Tip:
Always sketch the graph of the function when solving area problems to better visualize the boundaries and region you are calculating.

The problem asks for the area enclosed by the curve y = -x^2 - 3x + 4, the x-axis, and the lines x = -2 and the y-axis. What is the area of the enclosed region?

Learn how to calculate the area enclosed by the quadratic curve y = -x^2 - 3x + 4, the x-axis, and vertical lines x = -2 and x = 0 using definite integrals. This problem involves key calculus concepts, such as definite integrals and area calculations, and is suitable for students in Grades 10-12.

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