find the average value $f_{ave}$ of the function $f$ on the given interval. $f(x)=sqrt{x}, 0,4$

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{ave}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 0$, $b = 4$ and $f(x)=\sqrt{x}=x^{\frac{1}{2}}$. So $f_{ave}=\frac{1}{4-0}\int_{0}^{4}x^{\frac{1}{2}}dx$.

Step2: Integrate the function

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

Step3: Evaluate the definite integral

$\frac{2}{3}x^{\frac{3}{2}}\big|_{0}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}$. Since $(4)^{\frac{3}{2}}=(4^{\frac{1}{2}})^{3}=2^{3}=8$ and $(0)^{\frac{3}{2}} = 0$, we get $\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}=\frac{16}{3}$.

Step4: Calculate the average value

$f_{ave}=\frac{1}{4}\times\frac{16}{3}=\frac{4}{3}$.

Answer:

$\frac{4}{3}$