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.
