The curve with equation #y=x(3-x)^(1/2)# together with line segment OA.!) what are the coordinates of B and A; 2)what is the Area of the shaded region bounded by the line segment AO, x axis and the arc of AB curve?

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1 Answer
Narad T.
Jul 28, 2018

The coordinates of #B=(2,0)#. The point #A=(2,2)#. The area is

#=3.6u^2#

Explanation:

Calculation of the point #B#

Let

#y=xsqrt(3-x)=0#

Then,

#{(x=0 " this is the origin "O ),(x=3 " this is the point B"):}#

The coordinates of #B=(2,0)#

Calculation of the point #A#

Calculate the derivative of #y# according to the product rule

#dy/dx=(3-x)^(1/2)-x/(2(3-x)^(1/2))#

#=(2(3-x)-x)/(3-x)^(1/2)#

#=(6-3x)/(3-x)^(1/2)#

The maximum is when

#dy/dx=0#

That is

#6-3x=0#, #=>#, #x=2#

The point #A=(2,2)#

The equation of the line #OA# is

#y-0=1(x-0)#, #=>#, #y=x#

The area of the shaded region is

#A=int_0^2xdx+int_2^3xsqrt(3-x)dx#

#=int_2^3xsqrt(3-x)+int_0^2xdx#

#=I_1+I_2#

Calculation of #I_1# by substitution

Let #u=3-x#, #=>#, #du=-dx#

#x=3-u#

Therefore,

#I_1=int_2^3xsqrt(3-x)dx=int_1^0(3-u)sqrtu-du#

#=int_1^0u^(3/2)-3u^(1/2)du#

#=[2/5u^(5/2)-2u^(3/2)]_1^0#

#=(0)-(2/5*1^(5/2)-2*1^(3/2))#

#=8/5#

#=1.6#

#I_2=int_0^2xdx=[x^2/2]_0^2#

#=2#

Therefore, the area is

#A=1.6+2=3.6u^2#

graph{(y-xsqrt(3-x))(y-x)=0 [-0.96, 5.197, -0.3, 2.778]}

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