Question

The average value of the function F (T) =(t-4)^2 on [0,9] is

Answer 1

7

Explanation:

Average value of a function F(t) on interval [a,b]

is
`(1)/(b-a) \int_a^b F(x)dx`

So let's plug it in!

a=0

b=9

F(x)=(x-4)^2

To integrate (x-4)^2 just use power rule for integration.

So this what we get

`(1)/(9-0) ((x-4)^3)/(3)|_0^9`

`(1)/(9-0) [((9-4)^3)/(3)-((0-4)^3)/(3)]`

`(1)/(9) [(5^3)/(3)-((-4)^3)/(3)]`

`(1)/(9) \cdot (125+64)/(3)`

`(1)/(9) \cdot (189)/(3)`

7