Integration area??

Question

Find the area enclosed by the curves y=x(x-1) and y=x(2-x)

Answer 1

Curves intersect when :-
x² - x = 2x - x²
2x² - 3x = 0
x (2x - 3) = 0
x = 0 , x = 3 / 2
A(x) = ∫ x (2 - x) - x (x - 1) dx between x = 0 and x = 3 / 2
A(x) = ∫ 2x - x² - x² + x dx
A(x) = ∫ 3x - 2x² dx
A(x) = 3x² / 2 - 2 x³ / 3
A(x) = (3 / 2) (3 / 2)² - (2 / 3) (3 / 2)³
A(x) = (3 / 2)³ - (2 / 3) (27 / 8)
A(x) = 27 / 8 - 9 / 4
A(x) = 27 / 8 - 18 / 8
A(x) = 9 / 8 units ²

Answer 2

f(x) = x(x-1) and g(x) = x(2-x)

get the x-values of the intersection
f(x) = g(x) ........ x = 0 and 3/2

Note: f(1) = 0 and g(1) = 1, within the intersection, g is the upper function

Set-up of the area: ∫[0.. 3/2] {x(2-x) - x(x-1)} dx

d: