Answer
$\frac{31}{16}$
Work Step by Step
Step 1. Given the function $f(x)=x^3$ and the interval $[0,2]$, we can divide the interval into four parts of equal width of $\frac{1}{2}$ as shown in the figure. And we can find the middle points of each part as $x=\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}$
Step 2. We can find the height of each box (figure not to scale) as $f(\frac{1}{4})=(\frac{1}{4})^3, f(\frac{3}{4})=(\frac{3}{4})^3, f(\frac{5}{4})=(\frac{5}{4})^3, f(\frac{7}{4})=(\frac{7}{4})^3$
Step 3. We can approximate the area under the function over this interval as the sum of the four rectangles $A=\frac{1}{2}(f(\frac{1}{4})+f(\frac{3}{4})+f(\frac{5}{4})+f(\frac{7}{4}))=\frac{1}{2\cdot 4^3}(1^3+3^3+5^3+7^3)=\frac{496}{128}=\frac{31}{8}$
Step 4. The average value of the function over the interval is the area divided by the range; thus we have $\bar f_{[0,2]}=\frac{A}{2-0}=\frac{31}{16}$
