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*