The image contains the following integral expression:

$\int_{-2}^{0} \left( 3 - \frac{x}{2} \right) \, dx$

Let's solve this step by step:

Step 1: Break down the integral

We have:

$\int_{-2}^{0} \left( 3 - \frac{x}{2} \right) \, dx$

This is the sum of two simpler integrals:

$\int_{-2}^{0} 3 \, dx - \int_{-2}^{0} \frac{x}{2} \, dx$

Step 2: Solve each integral separately

First integral:

$\int_{-2}^{0} 3 \, dx = 3x \Big|_{-2}^{0} = 3(0) - 3(-2) = 6$

Second integral:

$\int_{-2}^{0} \frac{x}{2} \, dx = \frac{1}{2} \int_{-2}^{0} x \, dx$

The integral of $x$ is: $\frac{1}{2} \left( \frac{x^2}{2} \right) \Big|_{-2}^{0} = \frac{1}{2} \left( \frac{0^2}{2} - \frac{(-2)^2}{2} \right)$

$= \frac{1}{2} \left( 0 - \frac{4}{2} \right) = \frac{1}{2} \times (-2) = -1$

Step 3: Combine the results

Now, we add the two results:

$6 - 1 = 5$

So, the value of the integral is $5$.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you apply the Fundamental Theorem of Calculus to definite integrals?
2. What are some methods for solving more complex integrals?
3. How would the result change if the limits of the integral were reversed?
4. Can you explain the concept of an integral as the area under a curve?
5. How would the integral change if the function were multiplied by a constant?

Tip: When dealing with definite integrals, always evaluate the function at the limits and subtract the lower bound evaluation from the upper.