# How do you sketch the region enclosed by #y=1+sqrtx, Y=(3+x)/3# and find the area?

##### 1 Answer

#### Explanation:

Start by finding the intersection points of the two curves.

#1 + sqrt(x) = (3 + x)/3#

#3(1+ sqrt(x)) = 3 + x#

#3 + 3sqrt(x) =3 + x#

#3sqrt(x) = x+3 - 3#

#(3sqrtx)^2 = x^2#

#9x = x^2#

#0 = x^2-9x#

#0 = x(x-9)#

#x = 0 and 9#

Next, sketch a rudimentary graph of the curves to see which one lies above the other. You will find the graph of

Integrate:

#=int_0^9 sqrt(x) + 1 - (x + 3)/3 dx#

#=int_0^9 (3sqrtx + 3 - x - 3)/3dx#

#=int_0^9 (3sqrtx - x)/3 dx#

#=1/3int_0^9 3sqrtx -xdx#

#=1/3[2x^(3/2) - 1/2x^2]_0^9#

#=1/3(13.5)#

#=4.5# square units

Hopefully this helps