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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