Questions: What is the average value of 3-x^2 on [0,2]? 3 4 2 5/3

Solution Steps

We are given the function \( f(x) = 3 - x^2 \).

We need to find the integral of \( f(x) \) over the interval \([0, 2]\): \[ \int_{0}^{2} (3 - x^2) \, dx \] Calculating this integral, we find: \[ \int_{0}^{2} (3 - x^2) \, dx = \left[ 3x - \frac{x^3}{3} \right]_{0}^{2} = \left( 3(2) - \frac{(2)^3}{3} \right) - \left( 3(0) - \frac{(0)^3}{3} \right) = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

The length of the interval \([0, 2]\) is: \[ b - a = 2 - 0 = 2 \]

The average value of the function over the interval is given by: \[ \text{Average Value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx = \frac{1}{2} \cdot \frac{10}{3} = \frac{10}{6} = \frac{5}{3} \]

Final Answer