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.