Let's solve each part of the problem given the information provided.

Given Information:

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

We need to find the values of the following two integrals.

Problem 1

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

To find $\int_{-2}^{2} f(x) \, dx$, we can use the fact that the integral over $[-2, 5]$ is already given, and we can split this into two parts:

$\int_{-2}^{5} f(x) \, dx = \int_{-2}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx$

We know:

• $\int_{-2}^{5} f(x) \, dx = 16$
• $\int_{2}^{5} f(x) \, dx = 16$ (given directly in the problem)

Thus, $\int_{-2}^{2} f(x) \, dx = \int_{-2}^{5} f(x) \, dx - \int_{2}^{5} f(x) \, dx = 16 - 16 = 0$

So, $\int_{-2}^{2} f(x) \, dx = 0$

Problem 2

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

To evaluate this integral, we can separate it as follows:

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

Now, compute each part:

1. $\int_{-2}^{5} g(x) \, dx = 4$ (given directly).
2. For $\int_{-2}^{5} 3 \, dx$:
   • This is the integral of a constant, which becomes $3 \times$ the length of the interval $[-2, 5]$.
   • The length of the interval $[-2, 5]$ is $5 - (-2) = 7$.
   • So, $\int_{-2}^{5} 3 \, dx = 3 \times 7 = 21$.

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

Final Answers

1. $\int_{-2}^{2} f(x) \, dx = 0$
2. $\int_{-2}^{5} [g(x) + 3] \, dx = 25$

Would you like a more detailed explanation of any steps or concepts used here?

Related Questions:

1. What does it mean when we split an integral over two intervals?
2. How do we calculate the integral of a constant over an interval?
3. Why can we separate terms within an integral as shown in Problem 2?
4. What is the importance of understanding the limits of integration?
5. How would the answer change if $\int_{-2}^{5} f(x) \, dx$ were different?

Tip:

When evaluating integrals with known values over specific intervals, look for opportunities to split or combine intervals strategically to simplify calculations.