Math Problem Statement

Find the average value of the function f(x) = x^3 + 5x - 2 over the interval [2, 5].

Solution

To find the average value of the function f(x)=x3+5x2f(x) = x^3 + 5x - 2 over the interval [2,5][2, 5], we use the formula for the average value of a function over an interval [a,b][a, b]:

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

Here, f(x)=x3+5x2f(x) = x^3 + 5x - 2, a=2a = 2, and b=5b = 5. Let's compute this step by step.

Step 1: Find the integral of f(x)f(x)

We need to compute the indefinite integral of f(x)=x3+5x2f(x) = x^3 + 5x - 2:

f(x)dx=(x3+5x2)dx=x44+5x222x+C\int f(x) \, dx = \int (x^3 + 5x - 2) \, dx = \frac{x^4}{4} + \frac{5x^2}{2} - 2x + C

Step 2: Evaluate the definite integral

Now we evaluate the definite integral from 2 to 5:

25(x3+5x2)dx=[x44+5x222x]25\int_2^5 (x^3 + 5x - 2) \, dx = \left[ \frac{x^4}{4} + \frac{5x^2}{2} - 2x \right]_2^5

First, calculate at x=5x = 5:

544+5×5222×5=6254+125210=156.25+62.510=208.75\frac{5^4}{4} + \frac{5 \times 5^2}{2} - 2 \times 5 = \frac{625}{4} + \frac{125}{2} - 10 = 156.25 + 62.5 - 10 = 208.75

Next, calculate at x=2x = 2:

244+5×2222×2=164+2024=4+104=10\frac{2^4}{4} + \frac{5 \times 2^2}{2} - 2 \times 2 = \frac{16}{4} + \frac{20}{2} - 4 = 4 + 10 - 4 = 10

Now subtract:

208.7510=198.75208.75 - 10 = 198.75

Step 3: Compute the average value

Now divide by the length of the interval ba=52=3b - a = 5 - 2 = 3:

Average Value=13×198.75=66.25\text{Average Value} = \frac{1}{3} \times 198.75 = 66.25

Thus, the average value of the function over the interval [2,5][2, 5] is:

66.25\boxed{66.25}

Would you like any further details or have any other questions?

Here are five related questions to explore:

  1. How does the average value formula apply to functions that are negative on certain intervals?
  2. What is the geometric interpretation of the average value of a function over an interval?
  3. Can you find the average value of f(x)=x3+5x2f(x) = x^3 + 5x - 2 over a different interval, say [0,4][0, 4]?
  4. How does the presence of a cubic term in the function affect the shape of the graph?
  5. What role does definite integration play in finding the area under curves?

Tip: When working with definite integrals, always carefully evaluate the function at the upper and lower limits to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Average value of a function

Formulas

Average Value = (1 / (b - a)) * ∫[a to b] f(x) dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12