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To find the area between the curves, we need to set up the integral of the difference between the functions \( y = 2x^2 + 4 \) and \( y = 0 \) over the interval from \( x = -5 \) to \( x = 1 \).

The area \( A \) between the curves is given by the integral:
\[ A = \int_{-5}^{1} (2x^2 + 4 - 0) \, dx \]

Let's compute this step-by-step.

1. **Set up the integral:**
\[ A = \int_{-5}^{1} (2x^2 + 4) \, dx \]

2. **Integrate the function:**
\[ \int (2x^2 + 4) \, dx = \int 2x^2 \, dx + \int 4 \, dx \]

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

So, the antiderivative is:
\[ \frac{2}{3}x^3 + 4x \]

3. **Evaluate the definite integral:**
\[ A = \left[ \frac{2}{3}x^3 + 4x \right]_{-5}^{1} \]

Evaluate at the bounds \( x = 1 \) and \( x = -5 \):

For \( x = 1 \):
\[ \frac{2}{3}(1)^3 + 4(1) = \frac{2}{3} + 4 = \frac{2}{3} + \frac{12}{3} = \frac{14}{3} \]

For \( x = -5 \):
\[ \frac{2}{3}(-5)^3 + 4(-5) = \frac{2}{3}(-125) - 20 = -\frac{250}{3} - 20 = -\frac{250}{3} - \frac{60}{3} = -\frac{310}{3} \]

4. **Subtract the evaluated bounds:**
\[ A = \left( \frac{14}{3} \right) - \left( -\frac{310}{3} \right) \]
\[ A = \frac{14}{3} + \frac{310}{3} = \frac{324}{3} = 108 \]

Therefore, the area between the curves is:
\[ \boxed{108} \]

Would you like more details or have any questions? Here are some questions you might want to explore next:

1. How do you find the area between more complex curves?
2. What is the geometric interpretation of this integral?
3. Can you show the steps for solving a similar problem with different functions?
4. How does changing the limits of integration affect the area?
5. What are some real-world applications of finding the area between curves?

**Tip:** Always double-check your integration limits and functions to ensure the accuracy of the area calculation.

Learn how to calculate the area between curves using integration. Step-by-step solution for finding the area enclosed by the functions y = 2x^2 + 4 and y = 0 over the interval [-5, 1]. This problem demonstrates the use of integration to find geometric areas. Suitable for advanced high school students and anyone interested in calculus and geometric applications of integration.

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