What is the average value of 8-2x^3 on the interval [2,4]

To find the average value of the function \( f(x) = 8 - 2x^3 \) on the interval \([2, 4]\), 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, \( a = 2 \) and \( b = 4 \). So, we need to compute:<br /><br />\[<br />\text{Average value} = \frac{1}{4 - 2} \int_{2}^{4} (8 - 2x^3) \, dx<br />\]<br /><br />First, we find the integral:<br /><br />\[<br />\int_{2}^{4} (8 - 2x^3) \, dx<br />\]<br /><br />We can split this into two separate integrals:<br /><br />\[<br />\int_{2}^{4} 8 \, dx - \int_{2}^{4} 2x^3 \, dx<br />\]<br /><br />Evaluate each integral separately:<br /><br />\[<br />\int_{2}^{4} 8 \, dx = 8 \int_{2}^{4} 1 \, dx = 8 [x]_{2}^{4} = 8 (4 - 2) = 8 \cdot 2 = 16<br />\]<br /><br />\[<br />\int_{2}^{4} 2x^3 \, dx = 2 \int_{2}^{4} x^3 \, dx = 2 \left[ \frac{x^4}{4} \right]_{2}^{4} = 2 \left( \frac{4^4}{4} - \frac{2^4}{4} \right) = 2 \left( \frac{256}{4} - \frac{16}{4} \right) = 2 \left( 64 - 4 \right) = 2 \cdot 60 = 120<br />\]<br /><br />Now, combine the results:<br /><br />\[<br />\int_{2}^{4} (8 - 2x^3) \, dx = 16 - 120 = -104<br />\]<br /><br />Finally, compute the average value:<br /><br />\[<br />\text{Average value} = \frac{1}{4 - 2} \cdot (-104) = \frac{1}{2} \cdot (-104) = -52<br />\]<br /><br />Thus, the average value of \( 8 - 2x^3 \) on the interval \([2, 4]\) is \(\boxed{-52}\).