Question 421204
 Solve:

       {{{int(3x^2,dx, -2, 1)+int(3x,dx, -2, 1)-int( 6,dx, -2, 1)}}}  

       {{{ 3*int(x^2,dx, -2, 1)+3*int(x,dx,-2, 1)-int(6,dx, -2, 1)}}} 

        (x^3+(3/2)x^2- 6x)|(-2, 1)=[1^3+3/2*1^2-6*1 ]-[(-2)^3+3/2*  

        (-2)^2-6(-2)]=25/2 

 Answer: The area under the curve is 25/2 square units.