Answer:
2
Explanation:
The "average value of function f(x) on interval [a, b] is given by:
f(b) - f(a)
ave. value = ---------------
b - a
Here f(t)=(t-2)^2.
Thus, f(b) = (b - 2)^2. For b = 6, we get:
f(6) = 6^2 - 4(6) + 4, or f(6) = 36 - 24 + 4 = 16
For a = 0, we get:
f(0) = (0 - 2)^2 = 4
Plugging these results into the ave. value function shown above, we get:
16 - 4
ave. value = ------------ = 12/6 = 2
6 - 0
The average value of the function f(t)=(t-2)^2 on [0,6] is 2.