Question

Find the average value of f(x) =-t^2 + 10t + 50 on the interval [2,8]

Answer 1

find the area under the curve. The average value of f(x) will be the y-value of a rectangle with the same width as your region (i.e. 8-2= 6 units), and same area as your region.

in other words
\[ f_{avg}= \frac{1}{6} \int_2^8 -x^2 + 10x + 50 \ dx \]