Math Problem Statement

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

Solution

To find the average value of the function f(x)=5x5f(x) = 5x^5 on the interval [2,5][2, 5], we use the formula for the average value of a function:

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

Here:

  • a=2a = 2
  • b=5b = 5
  • f(x)=5x5f(x) = 5x^5

Step 1: Set up the formula

Average Value=152255x5dx\text{Average Value} = \frac{1}{5 - 2} \int_2^5 5x^5 \, dx Average Value=13255x5dx\text{Average Value} = \frac{1}{3} \int_2^5 5x^5 \, dx

Step 2: Compute the integral

5x5dx=5x66\int 5x^5 \, dx = \frac{5x^6}{6}

Now evaluate this from x=2x = 2 to x=5x = 5: 255x5dx=[5x66]25=5(56)65(26)6\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

56=15625,26=645^6 = 15625, \quad 2^6 = 64 5(56)6=5156256=781256,5(26)6=5646=3206\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} 255x5dx=7812563206=781253206=778056\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

Average Value=13778056=7780518\text{Average Value} = \frac{1}{3} \cdot \frac{77805}{6} = \frac{77805}{18}

Simplify if needed: 77805184322.5\frac{77805}{18} \approx 4322.5

Final Answer:

The average value of f(x)f(x) on [2,5][2, 5] is approximately 4322.5.

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


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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Average Value of a Function

Formulas

Average Value = (1 / (b - a)) ∫_a^b f(x) dx

Theorems

Mean Value Theorem for Integrals

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)