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Let's solve the problem step by step.

### Part (a): Find the average value of \( f(x) = 25 - x^2 \) on the interval \([0, 5]\).

The formula for the average value of a function \( f(x) \) on the interval \([a, b]\) is:

\[
\text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx
\]

In this case:
- \( f(x) = 25 - x^2 \)
- The interval is \([0, 5]\), so \( a = 0 \) and \( b = 5 \).

Thus, the average value is:

\[
\text{Average value} = \frac{1}{5 - 0} \int_0^5 (25 - x^2) \, dx
\]

We will now compute this integral.

\[
\int_0^5 (25 - x^2) \, dx = \int_0^5 25 \, dx - \int_0^5 x^2 \, dx
\]

The integrals are straightforward:
\[
\int_0^5 25 \, dx = 25x \Big|_0^5 = 25(5) - 25(0) = 125
\]
\[
\int_0^5 x^2 \, dx = \frac{x^3}{3} \Big|_0^5 = \frac{5^3}{3} - \frac{0^3}{3} = \frac{125}{3}
\]

Now, subtract the two results:
\[
\int_0^5 (25 - x^2) \, dx = 125 - \frac{125}{3} = \frac{375}{3} - \frac{125}{3} = \frac{250}{3}
\]

So the average value is:
\[
\text{Average value} = \frac{1}{5} \cdot \frac{250}{3} = \frac{250}{15} = \frac{50}{3} \approx 16.67
\]

### Part (b): Find a value \( c \) in the interval \([0, 5]\) such that \( f(c) \) is equal to the average value.

We need to solve the equation:
\[
f(c) = \frac{50}{3}
\]
Substituting \( f(c) = 25 - c^2 \):
\[
25 - c^2 = \frac{50}{3}
\]

First, subtract \( 25 \) from both sides:
\[
- c^2 = \frac{50}{3} - 25 = \frac{50}{3} - \frac{75}{3} = -\frac{25}{3}
\]

Multiply by \( -1 \) to get:
\[
c^2 = \frac{25}{3}
\]

Now, take the square root:
\[
c = \pm \sqrt{\frac{25}{3}} = \pm \frac{5}{\sqrt{3}} = \pm \frac{5\sqrt{3}}{3}
\]

Since \( c \) must be in the interval \([0, 5]\), we take the positive root:
\[
c = \frac{5\sqrt{3}}{3} \approx 2.89
\]

### Final Answers:
- (a) The average value of \( f(x) \) on the interval \([0, 5]\) is \( \frac{50}{3} \) or approximately 16.67.
- (b) The value \( c \) such that \( f(c) \) equals the average value is \( c = \frac{5\sqrt{3}}{3} \) or approximately 2.89.

Would you like further details on any part of the solution? Here are some follow-up questions for deeper understanding:

1. How does the average value formula relate to the definite integral?
2. What is the significance of finding a point where \( f(c) \) equals the average value?
3. Can you apply the same process to a different function on the same interval?
4. What happens to the average value if the interval is changed?
5. What are some common mistakes when calculating definite integrals?

**Tip:** Always double-check your integral bounds and limits when performing definite integration to avoid small calculation errors!

Find the average value of f(x) = 25 - x^2 on the interval [0, 5]. Find a value c in the interval [0,5] such that f(c) is equal to the average value.

This problem involves finding the average value of the function f(x) = 25 - x^2 on the interval [0, 5], using integral calculus. It also requires determining a value c such that f(c) equals the average value. The problem covers key concepts from integral calculus and the average value theorem, suitable for students in Grades 11-12 or early college.

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