Our first step is to find the interval over which we have to integrate. This is accomplished by setting the 2 functions equal to each other. And then solve for x.
#1+sqrt(x)=1+x/3#, subtract #1# from both sides
#sqrt(x)=x/3#, square both sides
#x=(x^2)/9#
#x-(x^2)/9=0#
#x(1-x/9)=0#
Set each factor equal to 0.
#x=0-># lower bound
#1-x/9=0#
#-x/9=-1#
#x=9-># upper bound
#[0,9] -># interval
We now need to figure out which function is greater over this interval. To do this we substitute in a value between 0 and 9. Lets us an #x# value of #1#.
#y=1+sqrt(1)=1+1=2 -># Larger function
#y=1+1/3=4/3=1.333#
#int_0^9 1+sqrt(x)-(1+x/3)dx#
#int_0^9 1+sqrt(x)-1-x/3dx#
#int_0^9 sqrt(x)-x/3dx#
#int_0^9 x^(1/2)-x/3dx#
#[x^(3/2)/(3/2)-x^2/(3*2)]_0^9#
#[(9)^(3/2)/(3/2)-(9)^2/(3*2)-((0)^(3/2)/(3/2)-(0)^2/(3*2))]#
#[(9)^(3/2)/(3/2)-(9)^2/(3*2)]#
#[sqrt(9^3)/(3/2)-81/6]#
#[sqrt(9^3)*(2/3)-81/6]#
#[sqrt((3^2)^3)*(2/3)-81/6]#
#[sqrt((3^6))*(2/3)-81/6]#
#[(3^3)*(2/3)-81/6]#
#[(3^2)*(2)-81/6]#
#[(9)*(2)-81/6]#
#[18-27/2]#
#[36/2-27/2]#
#[9/2]=4.5 -># Solution
