Let's solve these integrals step-by-step, based on the information provided.

Given Information

1. $\int_{-2}^{5} f'(x) \, dx = -5$
2. $\int_{2}^{5} f(x) \, dx = 16$
3. $\int_{-2}^{5} g(x) \, dx = 4$

We need to find:

1. $\int_{-2}^{2} f(x) \, dx$
2. $\int_{-2}^{5} [g(x) + 3] \, dx$

Solution

Part 1: $\int_{-2}^{2} f(x) \, dx$

We can use the information about $\int_{2}^{5} f(x) \, dx$ and the property that: $\int_{-2}^{5} f(x) \, dx = \int_{-2}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx$ However, $\int_{-2}^{5} f(x) \, dx$ is not given directly. We are only given $\int_{-2}^{5} f'(x) \, dx = -5$.

Using the Fundamental Theorem of Calculus: $\int_{-2}^{5} f'(x) \, dx = f(5) - f(-2) = -5$

Since we don't have the values of $f(5)$ and $f(-2)$, we cannot proceed with this integral based on the given data.

Part 2: $\int_{-2}^{5} [g(x) + 3] \, dx$

Using the linearity of integration, we can separate the integral as follows: $\int_{-2}^{5} [g(x) + 3] \, dx = \int_{-2}^{5} g(x) \, dx + \int_{-2}^{5} 3 \, dx$

We already know that $\int_{-2}^{5} g(x) \, dx = 4$.

Now, calculate $\int_{-2}^{5} 3 \, dx$: $\int_{-2}^{5} 3 \, dx = 3 \times (5 - (-2)) = 3 \times 7 = 21$

Thus: $\int_{-2}^{5} [g(x) + 3] \, dx = 4 + 21 = 25$

Answers

1. $\int_{-2}^{2} f(x) \, dx$ (Cannot be determined with the given information)
2. $\int_{-2}^{5} [g(x) + 3] \, dx = 25$

Would you like further details on any of these steps, or do you have any other questions?

Here are five related questions for further practice:

1. How would you solve if given $\int_{a}^{b} f(x) \, dx$ and $\int_{c}^{d} f(x) \, dx$ but needed $\int_{a}^{d} f(x) \, dx$?
2. What is the significance of the Fundamental Theorem of Calculus in solving these types of integrals?
3. How can you determine $\int_{-2}^{5} f(x) \, dx$ if only partial interval values are known?
4. How would you evaluate $\int_{-2}^{5} h(x) \, dx$ if given multiple functions to combine within the integral?
5. Can we always separate a function's integral into parts like $\int [g(x) + c] \, dx$?

Tip: When breaking down integrals over intervals, always consider any given values across sub-intervals or derivative relations.