What is the average value of (1)/(x) on the interval 4leqslant xleqslant 8

To find the average value of the function \( f(x) = \frac{1}{x} \) on the interval \([4, 8]\), 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 />In this case, \( f(x) = \frac{1}{x} \), \( a = 4 \), and \( b = 8 \). So, we need to compute the integral:<br /><br />\[<br />\int_{4}^{8} \frac{1}{x} \, dx<br />\]<br /><br />The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). Therefore,<br /><br />\[<br />\int_{4}^{8} \frac{1}{x} \, dx = \left. \ln|x| \right|_{4}^{8} = \ln(8) - \ln(4) = \ln\left(\frac{8}{4}\right) = \ln(2)<br />\]<br /><br />Now, we can find the average value:<br /><br />\[<br />\text{Average value} = \frac{1}{8-4} \cdot \ln(2) = \frac{1}{4} \cdot \ln(2)<br />\]<br /><br />Thus, the average value of \(\frac{1}{x}\) on the interval \([4, 8]\) is:<br /><br />\[<br />\boxed{\frac{1}{4} \ln(2)}<br />\]