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 \([-1, 3]\), we use the formula for the average value of a continuous function over an interval \([a, b]\):<br /><br />\[<br />\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx<br />\]<br /><br />Here, \( a = -1 \) and \( b = 3 \). So, we need to compute:<br /><br />\[<br />\text{Average value} = \frac{1}{3 - (-1)} \int_{-1}^{3} (14 - 6x^2) \, dx<br />\]<br /><br />First, simplify the interval length:<br /><br />\[<br />3 - (-1) = 4<br />\]<br /><br />Next, we need to evaluate the integral:<br /><br />\[<br />\int_{-1}^{3} (14 - 6x^2) \, dx<br />\]<br /><br />We can split this integral into two parts:<br /><br />\[<br />\int_{-1}^{3} 14 \, dx - \int_{-1}^{3} 6x^2 \, dx<br />\]<br /><br />Evaluate each integral separately:<br /><br />1. \(\int_{-1}^{3} 14 \, dx = 14 \int_{-1}^{3} 1 \, dx = 14 [x]_{-1}^{3} = 14 (3 - (-1)) = 14 \cdot 4 = 56\)<br /><br />2. \(\int_{-1}^{3} 6x^2 \, dx = 6 \int_{-1}^{3} x^2 \, dx = 6 \left[ \frac{x^3}{3} \right]_{-1}^{3} = 6 \left( \frac{3^3}{3} - \frac{(-1)^3}{3} \right) = 6 \left( \frac{27}{3} - \frac{-1}{3} \right) = 6 \left( 9 + \frac{1}{3} \right) = 6 \cdot \frac{28}{3} = 56\)<br /><br />Now, combine the results of the integrals:<br /><br />\[<br />\int_{-1}^{3} (14 - 6x^2) \, dx = 56 - 56 = 0<br />\]<br /><br />Finally, divide by the length of the interval:<br /><br />\[<br />\text{Average value} = \frac{0}{4} = 0<br />\]<br /><br />Thus, the average value of \( 14 - 6x^2 \) on the interval \([-1, 3]\) is:<br /><br />\[<br />\boxed{0}<br />\]