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
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