Answer
$1$
Work Step by Step
Step 1. Given the function $f(x)=\frac{1}{2}+sin^2(\pi t)$ 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 $t=\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}$
Step 2. We can find the height of each box (figure not to scale) as $f(\frac{1}{4})=\frac{1}{2}+sin^2(\pi/4)=1, f(\frac{3}{4})=\frac{1}{2}+sin^2(3\pi/4)=1, f(\frac{5}{4})=\frac{1}{2}+sin^2(5\pi/4)=1, f(\frac{7}{4})=\frac{1}{2}+sin^2(7\pi/4)=1$
Step 3. We can approximate the area under the function over this interval as the sum of the four rectangles $A=\frac{1}{2}(f(\frac{1}{4})+f(\frac{3}{4})+f(\frac{5}{4})+f(\frac{7}{4}))=\frac{1}{2}(1+1+1+1)=2$
Step 4. The average value of the function over the interval is the area divided by the interval range; thus we have $\bar f_{[0,2]}=\frac{A}{2-0}=1$
