Let's solve the questions step by step based on the problem provided:

Given Information:

• $\int_0^5 f(x) \, dx = 10$
• $f(x)$ is continuous and non-negative on $[0, 5]$.

You are tasked to find:

1. $\int_{-5}^5 f(x) \, dx$
2. $\int_{-5}^5 |f(x)| \, dx$

Under the following conditions:

1. $f(x)$ is an even function.
2. $f(x)$ is an odd function.

1. If $f(x)$ is an even function:

For even functions, $f(-x) = f(x)$.

1. Finding $\int_{-5}^5 f(x) \, dx$:

   • Due to symmetry: $\int_{-5}^5 f(x) \, dx = 2 \int_0^5 f(x) \, dx.$
   • Substituting the given value: $\int_{-5}^5 f(x) \, dx = 2 \cdot 10 = 20.$

2. Finding $\int_{-5}^5 |f(x)| \, dx$:

   • Since $f(x) \geq 0$ (non-negative), $|f(x)| = f(x)$. $\int_{-5}^5 |f(x)| \, dx = \int_{-5}^5 f(x) \, dx.$
   • Thus: $\int_{-5}^5 |f(x)| \, dx = 20.$

2. If $f(x)$ is an odd function:

For odd functions, $f(-x) = -f(x)$.

1. Finding $\int_{-5}^5 f(x) \, dx$:

   • Due to symmetry, the integral of an odd function over symmetric limits is 0: $\int_{-5}^5 f(x) \, dx = 0.$

2. Finding $\int_{-5}^5 |f(x)| \, dx$:

   • The absolute value of $f(x)$ (an odd function) makes it an even function: $|f(x)| = -f(x) \text{ for } x < 0 \quad \text{and} \quad |f(x)| = f(x) \text{ for } x \geq 0.$
   • Thus: $\int_{-5}^5 |f(x)| \, dx = 2 \int_0^5 f(x) \, dx.$
   • Substituting the given value: $\int_{-5}^5 |f(x)| \, dx = 2 \cdot 10 = 20.$

Final Answers:

1. When $f(x)$ is even:

   • $\int_{-5}^5 f(x) \, dx = 20$
   • $\int_{-5}^5 |f(x)| \, dx = 20$

2. When $f(x)$ is odd:

   • $\int_{-5}^5 f(x) \, dx = 0$
   • $\int_{-5}^5 |f(x)| \, dx = 20$

Do you need any clarifications or further breakdowns?

Here are some related questions:

1. How do odd and even functions affect integrals over symmetric intervals?
2. Why does the absolute value affect odd functions but not even functions?
3. Can we compute $\int_{a}^b |f(x)| \, dx$ directly without symmetry arguments?
4. What happens if $f(x)$ is neither odd nor even?
5. How does the non-negativity of $f(x)$ simplify this problem?

Tip: Always use symmetry properties of functions to simplify definite integrals over symmetric intervals.