Question

Question #03038

Answer 1

`f_(ave)=1`
`c=6`
`c=8`

Explanation:

The average value of a function is found by:

`f_(ave)=1/(b-a)*int_a^bf(x)dx`

on some interval `[a,b]`

We have `f(x)=(x-7)^2,a=6,` and `b=9`.

Therefore:

`f_(ave)=1/3int_6^9(x-7)^2dx`

We can use a basic substitution to solve, where `u=x-7`

`=1/3int_(-1)^2u^2du`

`=>=1/3*1/3u^3]_(-1)^2`

`=1/9(8+1)`

`=1`

To find `c` such that `f_(ave)=f(c)`, we can use the mean value theorem for integrals, which says that if you have a function which is continuous on a closed interval `[a,b]`, then there is some number `c` in `[a,b]` such that `f(c)=f_(ave)`.

We found `f_(ave)=1,` so `f(c)=1`.

`=>(c-7)^2=1`

`=>c^2-14c+49=1`

`=>c^2-14c+48=0`

We find that `c=8` or `c=6`. Both of these numbers are contained within the given interval.

`c=6: f(6)=(6-7)^2=1`

`c=8: f(8)=(8-7)^2=1`

The area of the rectangle is given by `(b-a)f(c)` where `(b-a)` is the width and `f(c)` is the height. This gives the rectangle a width of `3` and a height of `1`. You would therefore want the top right graph.