Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 5

Given the curve equations ,they are
y_1 = 3(x^3 - x) -----(1)
y_2 = 0 -----(2)
to get the boundaries or the intersecting points of the functions we have to equate the functions .
y_1=y_2
=> 3(x^3 - x)= 0
=> (x^3 - x)=0
=> x(x^2-1)=0
=> x=0 or x=+-1
so,
so,
the Area =int_-1^0 3(x^3)-x) -0 dx + int_0 ^1 0-(3x^3-x) dx
= int_(-1) ^0 (3x^3-3x) -0 dx +int_(0) ^1 0-(3x^3-3x) dx
= [(3x^4)/4 -3/2 x^2]_(-1) ^0 +[-(3x^4)/4 +3/2 x^2]_(0) ^1
=[0]-[3/4 - 3/2] +[-3/4+3/2]-[0]
=-3/4 +3/2-3/4+3/2
= -3/2 +3
=3/2 = 1.5 is the area of the region between the curves