# `y_1 = 3(x^3 - x) , y_2 = 0` Set up the definite integral that gives the area of the region 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...

## See This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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`

`=-[3/4 - 3/2] +[-3/4+3/2]-`

=`-3/4 +3/2-3/4+3/2`

= `-3/2 +3`

=`3/2 = 1.5` is the area of the region between the curves

Approved by eNotes Editorial Team