Math Problem Statement

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.

Solution

Let's solve the problem step by step.

Part (a): Find the average value of f(x)=25x2f(x) = 25 - x^2 on the interval [0,5][0, 5].

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

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

In this case:

  • f(x)=25x2f(x) = 25 - x^2
  • The interval is [0,5][0, 5], so a=0a = 0 and b=5b = 5.

Thus, the average value is:

Average value=15005(25x2)dx\text{Average value} = \frac{1}{5 - 0} \int_0^5 (25 - x^2) \, dx

We will now compute this integral.

05(25x2)dx=0525dx05x2dx\int_0^5 (25 - x^2) \, dx = \int_0^5 25 \, dx - \int_0^5 x^2 \, dx

The integrals are straightforward: 0525dx=25x05=25(5)25(0)=125\int_0^5 25 \, dx = 25x \Big|_0^5 = 25(5) - 25(0) = 125 05x2dx=x3305=533033=1253\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: 05(25x2)dx=1251253=37531253=2503\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: Average value=152503=25015=50316.67\text{Average value} = \frac{1}{5} \cdot \frac{250}{3} = \frac{250}{15} = \frac{50}{3} \approx 16.67

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

We need to solve the equation: f(c)=503f(c) = \frac{50}{3} Substituting f(c)=25c2f(c) = 25 - c^2: 25c2=50325 - c^2 = \frac{50}{3}

First, subtract 2525 from both sides: c2=50325=503753=253- c^2 = \frac{50}{3} - 25 = \frac{50}{3} - \frac{75}{3} = -\frac{25}{3}

Multiply by 1-1 to get: c2=253c^2 = \frac{25}{3}

Now, take the square root: c=±253=±53=±533c = \pm \sqrt{\frac{25}{3}} = \pm \frac{5}{\sqrt{3}} = \pm \frac{5\sqrt{3}}{3}

Since cc must be in the interval [0,5][0, 5], we take the positive root: c=5332.89c = \frac{5\sqrt{3}}{3} \approx 2.89

Final Answers:

  • (a) The average value of f(x)f(x) on the interval [0,5][0, 5] is 503\frac{50}{3} or approximately 16.67.
  • (b) The value cc such that f(c)f(c) equals the average value is c=533c = \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)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!

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Average Value of a Function

Formulas

Average value of a function: (1 / (b - a)) * integral from a to b of f(x) dx
Integral of x^2: (x^3)/3

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, Early College