three regions R1 R2 and R3 are formed when the curves of y2s

Solution

with reference to my previous diagram,
y = 3 - x and y = x(x-3), solving simultaneously
3 - x = x(x-3)
x(x-3) + (x-3) = 0 => (x+1)(x-3) = 0 => x = -1 and x = 3
y = 4 for x = -1
y = 0 for x = 3
so points are A(-1,4) and B(3,0)

y = 2x and y = 3 - x, solving simultaneously

y^2 = 4x , y = 3 - x

(3-x)^2 = 4x

=> x^2 - 6x +9 = 4x

=> x^2 - 10x + 9 = 0

=> (x-9)(x-1) = 0

x = 9, x = 1

but x should not be 9, y = 29 = 6, y = 3 - 9 = -6

6 not equal to -6

hence x = 9 is rejected

y = 2 for x = 1

hence point is C(1,2)

y = 2x and y = x(x-3), solving simultaneously

y^2 = 4x

4x = x^2(x-3)^2

4 = x(x-3)^2

x = 4, satisfies above

y = 4 for x = 4

D(4,4)

now we got all points of intersections, ready to calculate areas R1, R2 and R3

area R1 = integ (3-x) dx from A to C - integ 2c from A to O(origin) - integ x(x-3)dx from O to C

= (3x-x^2/2) from -1 to 1 - 4/3x^(3/2) from 0 to 1 - (x^3/3 - 3x^2/2) from -1 to 0

= 2.5 - (-3 -1/2) -[ 4/3 + 3/2 + 1/3]

= 2.5 + 3.5 - [1.5 + 5/3] = 9/2 - 5/3 = 17/6 units

R2 = integ 2x dx from C to D - integ (3-x) dx from C to B - integ x(x-3) dx from B to D

4/3x^3/2 from 1 to 4 - (3x - x^2/2) from 1 to 3 - (x^3/3 - 3x^2/2) from 3 to 4

= 32/3 - (2) - (64/3 - 24 - 9 + 27/2)

= 6.83333 units

R3 = integ 2x dx from O to C + integ (3-x) dx from C to B + integ x(x-3) dx from O to B

= 4/3x^3/2 from 0 to 1 + (3x-x^2/2) from 1 to 3 + (x^3/3 - 3x^2/2) from 0 to 3

= 4/3 + 2 + 9 - 9/2

= 10/3 + 9/2

= 47/6 units