Find the average value of the function f(t)=(t-2)^2 on [0,6]

2

Step-by-step 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.

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope

757764

step-by-step explanation:

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