Question:

The average value of function \( f(x) = \sqrt{9 - x^2} \) on \([-3, 3]\), rounded off to TWO decimal places, is ............ 
 

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To calculate the average value of a function, use the formula \( \frac{1}{b - a} \int_a^b f(x) \, dx \), and recognize geometric shapes when possible (like the semicircle in this case).
Updated On: Dec 4, 2025
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Correct Answer: 2.3 - 2.4

Solution and Explanation

Step 1: Formula for the average value of a function.
The average value of a function \( f(x) \) on the interval \( [a, b] \) is given by the formula: \[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx. \]

Step 2: Applying the formula to the given function.
For \( f(x) = \sqrt{9 - x^2} \) on \( [-3, 3] \), the average value is: \[ \frac{1}{3 - (-3)} \int_{-3}^{3} \sqrt{9 - x^2} \, dx = \frac{1}{6} \int_{-3}^{3} \sqrt{9 - x^2} \, dx. \]

Step 3: Recognizing the integral.
The integral \( \int_{-3}^{3} \sqrt{9 - x^2} \, dx \) is the area of a semicircle with radius 3, which is: \[ \text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3)^2 = \frac{9\pi}{2}. \]

Step 4: Calculating the average value.
Substituting this into the formula for the average value: \[ \text{Average value} = \frac{1}{6} \times \frac{9\pi}{2} = \frac{3\pi}{4} \approx 2.356. \]

Step 5: Conclusion.
The average value of the function is \( \boxed{2.36} \).

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