Let R be the region in the first quadrant bounded above by the graph of #y=(6x+4)^(1/2)# the line #y=2x# and the y axis, how do you find the area of region R?

1 Answer
Nov 28, 2017

Answer:

The area is #=20/9u^2#

Explanation:

The point of intersection of the line #y=2x# with the curve #y=sqrt(6x+4)# is

#sqrt(6x+4)=2x#

#6x+4=4x^2#

#2x^2-3x-2=0#

#x=(3+-sqrt(9+16))/(4)=(3+-5)/(4)#

#x=2# or #x=-1/2#

#intsqrt(6x+4)=(6x+4)^(3/2)/(3/2*6)=1/9(6x+4)^(3/2)+C#

#int2xdx=2*x^2/2+C#

Therefore,

The area is

#A=int_0^2(sqrt(6x+4)-2x)dx#

#=[1/9(6x+4)^(3/2)-x^2]_0^2#

#=(1/9(16)^(3/2)-4)-(1/9*4^(3/2))#

#=28/9-8/9#

#=20/9u^2#

graph{(y-sqrt(6x+4))(y-2x)=0 [-3.9, 10.146, -0.91, 6.11]}