How do you find the area of the region bounded by the curves #y=1+sqrt(x)# and #y=1+x/3# ?

Question

Emma Aldridge · Accepted Answer

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

Emilia Aguilar · Answer

undefined To find the area of the region bounded by the curves $y = 1 + \sqrt{x}$ and $y = 1 + \frac{x}{3}$, you need to find the points of intersection of the two curves first. Set the two equations equal to each other and solve for $x$. Once you have the $x$-coordinates of the points of intersection, integrate the difference between the curves from the smaller $x$-value to the larger $x$-value to find the area. The integral should be:

$$A = \int_{x_1}^{x_2} (1 + \sqrt{x} - (1 + \frac{x}{3})) \, dx$$

where $x_1$ and $x_2$ are the $x$-coordinates of the points of intersection.