Let
#G_1 : y = 4^x=2^(2x)#
#G_2 : y = 2(2^x)#
#G_3 : y = 2^x+6#
#G_2 and G_3# meet at #A ( log 6/log 2, 12 )= (2.585, 12 )#, nearly,
using #2(2^x)-2^x+6#
#G_3 and G_1# meet ( after discarding negative #2^x#)
at #B (log 3/log 2, 9)=(1.585, 9 )#, nearly,
using #4^x=(2^x)^2=2^x+6#.
#G_1 and G_2# meet at C ( 1, 4 ), using #4^x=2(2^x)#.
Note that B is between farther A and nearer C, in the x-direction and is above the curve #G_2 (AC)#.
Now, the area under reference is the area enclosed by the curved
triangle ABC.
The vertical x = #x_B#= 1.585 divides the area into two parts
Area = # int (y_(G_1)-y_(G_2)) dx#, from x = #x_C=1# to #x_B=1.585#
#+ int (y_(G_3)-y_(G_2)) dx#, from x = #x_B=1.585# to #x_A=2.585#
#= int (4^x-2(2^x) dx#, for x from 1 to 1.585
#+ int (2^x+6-2(2^x)) dx#, for x from 1.585 to 2.585
#=[ 4^x/ln 4-2(2^x/ln 2]#, between x = 1 and 1.585
#+[ 2^x/ln2+6x-2(2^x)/ln2]]#, between x = 1.585 and 2.585
#=(3.4924-2.8854)+(6.8536-5.1818)#
#=2.279#, nearly..
I would review my answer, a little later, for probable errors.
