Evaluate the area between the graphs y1=x and y2= (x+3)x(x-3) ?

Firstly, find the points of intersection of the y₁ = x and y₂ = (x + 3) x (x - 3). Two simultaneous equations are :
y = x ...... [1]
y = (x + 3) x (x - 3) ...... [2]

[2] = [1] :
(x + 3) x (x - 3) = x
(x + 3) x (x - 3) - x = 0
x [ (x + 3)(x - 3) - 1] = 0
x (x² - 9 - 1) = 0
x (x² - 10) = 0
x = 0 or x = √10 or x = -√10

The two curves meet at x = -√10, x = 0 and x = √10

When x is in the interval (-√10, 0), y₂ lies above y₁.
When x is in the interval (0, √10), y₁ lies above y₂.

Area between the two given curves
= ∫_(-√10 to 0) [(x + 3) x (x - 3) - x] dx + ∫_(-√10 to 0) [x - (x + 3) x (x - 3)] dx
= ∫_(-√10 to 0) (x³ - 10x) dx + ∫_(-√10 to 0) (10x - x³) dx
= [(1/4)x⁴ - 5x²]_(-√10 to 0) + [5x² - (1/4)x⁴]_(0 to √10)
= 0 - [(1/4)(-√10)⁴ - 5(-√10)²] + [5(√10)² - (1/4)(√10)⁴] - 0
= 50 (sq. units)