Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 35

y=3x^2-2x , y=x^3-3x+4
Refer the attached image. Graph of y=3x^2-2x is plotted in red color and graph of y=x^3-3x+4 is plotted in blue color.
From the graph, the x-coordinates of the intersection of curves are x ~~ -1.1 , x ~~ 1.25 , x ~~ 2.875.
Area of the region bounded by the curves A =int_(-1.1)^(1.25)((x^3-3x+4)-(3x^2-2x))dx+int_1.25^2.875((3x^2-2x)-(x^3-3x+4))dx
=int_(-1.1)^1.25(x^3-3x+4-3x^2+2x)dx+int_1.25^2.875(3x^2-2x-x^3+3x-4)dx
=int_(-1.1)^1.25(x^3-3x^2-x+4)dx+int_1.25^2.875(-x^3+3x^2+x-4)dx
=[x^4/4-3x^3/3-x^2/2+4x]_(-1.1)^1.25+[-x^4/4+3x^3/3+x^2/2-4x]_1.25^2.875
=[x^4/4-x^3-x^2/2+4x]_(-1.1)^1.25+[-x^4/4+x^3+x^2/2-4x]_1.25^2.875
=(1.25^4/4-1.25^3-1.25^2/2+4*1.25)-((-1.1)^4/4-(-1.1)^3-(-1.1)^2/2+4(-1.1))+(-(2.875)^4/4+2.875^3+2.875^2/2-4(2.875))-(-(1.25)^4/4+1.25^3+1.25^2/2-4(1.25)) =(2.875976563)-(-3.307975)+(-0.683654785)-(-2.875976563) =8.37627334