Nhat is the average value of 14-6x^2 on the interval [-1,3] square

To find the average value of the function \( f(x) = 14 - 6x^2 \) on the interval , we use the formula for the average value of a continuous function over an interval : Here, and . So, we need to compute: First, simplify the interval length: Next, we need to evaluate the integral: We can split this integral into two parts: Evaluate each integral separately:1. \(\int_{-1}^{3} 14 \, dx = 14 \int_{-1}^{3} 1 \, dx = 14 [x]_{-1}^{3} = 14 (3 - (-1)) = 14 \cdot 4 = 56\)2. \(\int_{-1}^{3} 6x^2 \, dx = 6 \int_{-1}^{3} x^2 \, dx = 6 \left[ \frac{x^3}{3} \right]_{-1}^{3} = 6 \left( \frac{3^3}{3} - \frac{(-1)^3}{3} \right) = 6 \left( \frac{27}{3} - \frac{-1}{3} \right) = 6 \left( 9 + \frac{1}{3} \right) = 6 \cdot \frac{28}{3} = 56\)Now, combine the results of the integrals: Finally, divide by the length of the interval: Thus, the average value of on the interval is: