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To find the average value of the function \( f(x) = 5x^5 \) on the interval \( [2, 5] \), we use the formula for the average value of a function:

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

Here:
- \( a = 2 \)
- \( b = 5 \)
- \( f(x) = 5x^5 \)

### Step 1: Set up the formula
\[
\text{Average Value} = \frac{1}{5 - 2} \int_2^5 5x^5 \, dx
\]
\[
\text{Average Value} = \frac{1}{3} \int_2^5 5x^5 \, dx
\]

### Step 2: Compute the integral
\[
\int 5x^5 \, dx = \frac{5x^6}{6}
\]

Now evaluate this from \( x = 2 \) to \( x = 5 \):
\[
\int_2^5 5x^5 \, dx = \left[ \frac{5x^6}{6} \right]_2^5 = \frac{5(5^6)}{6} - \frac{5(2^6)}{6}
\]

### Step 3: Simplify the terms
\[
5^6 = 15625, \quad 2^6 = 64
\]
\[
\frac{5(5^6)}{6} = \frac{5 \cdot 15625}{6} = \frac{78125}{6}, \quad \frac{5(2^6)}{6} = \frac{5 \cdot 64}{6} = \frac{320}{6}
\]
\[
\int_2^5 5x^5 \, dx = \frac{78125}{6} - \frac{320}{6} = \frac{78125 - 320}{6} = \frac{77805}{6}
\]

### Step 4: Compute the average value
\[
\text{Average Value} = \frac{1}{3} \cdot \frac{77805}{6} = \frac{77805}{18}
\]

Simplify if needed:
\[
\frac{77805}{18} \approx 4322.5
\]

### Final Answer:
The average value of \( f(x) \) on \( [2, 5] \) is approximately **4322.5**.

Let me know if you'd like further details or a breakdown of any step!

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**Questions to explore:**
1. How does the formula for the average value of a function relate to the Mean Value Theorem for integrals?
2. What are some practical applications of finding the average value of a function?
3. How would the average value change if the interval were doubled?
4. How can we approximate the average value numerically if the integral is too complex to compute by hand?
5. Why is the average value dependent on the interval's width?

**Tip:** Always verify your limits of integration and simplify terms carefully to avoid minor calculation errors.

Find the average value of the function f(x) = 5x^5 on the interval 2 ≤ x ≤ 5.

Calculate the average value of the function f(x) = 5x^5 over the interval [2, 5] using calculus. This problem demonstrates the use of definite integrals and the formula for the average value of a function, suitable for advanced high school students.

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