`y = 3x^2 - 2x, y = x^3 - 3x + 4` Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find...

`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`

`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`