compute the average of $h(x)=\frac{ln(x)}{x}$ on the interval $1,4$.

Explanation:

Step1: Recall average - value formula

The average value of a function $y = h(x)$ on the interval $[a,b]$ is given by $\bar{h}=\frac{1}{b - a}\int_{a}^{b}h(x)dx$. Here, $a = 1$, $b = 4$ and $h(x)=\frac{\ln(x)}{x}$, so $\bar{h}=\frac{1}{4 - 1}\int_{1}^{4}\frac{\ln(x)}{x}dx=\frac{1}{3}\int_{1}^{4}\frac{\ln(x)}{x}dx$.

Step2: Use substitution

Let $u=\ln(x)$, then $du=\frac{1}{x}dx$. When $x = 1$, $u=\ln(1)=0$; when $x = 4$, $u=\ln(4)$. The integral $\frac{1}{3}\int_{1}^{4}\frac{\ln(x)}{x}dx$ becomes $\frac{1}{3}\int_{0}^{\ln(4)}u du$.

Step3: Integrate $u$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n = 1$, $\int udu=\frac{u^{2}}{2}+C$. So, $\frac{1}{3}\int_{0}^{\ln(4)}u du=\frac{1}{3}\left[\frac{u^{2}}{2}\right]_{0}^{\ln(4)}$.

Step4: Evaluate the definite integral

$\frac{1}{3}\left[\frac{u^{2}}{2}\right]_{0}^{\ln(4)}=\frac{1}{3}\times\frac{(\ln(4))^{2}-0^{2}}{2}=\frac{(\ln(4))^{2}}{6}=\frac{(2\ln(2))^{2}}{6}=\frac{4(\ln(2))^{2}}{6}=\frac{2(\ln(2))^{2}}{3}$.

Answer:

$\frac{2(\ln(2))^{2}}{3}$