Derek F. answered • 02/02/16

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The area below the x-axis between roots x=2 and x =3 is

∫

^{3}_{2}x^3-6x^2+11x-6 dx = (x^4)/4-2x^3+(11x^2)/2-6x |^{3}_{2}= [((3)^4)/4-2(3)^3+(11(3)^2)/2-6(3)] - [((2)^4)/4-2(2)^3+(11(2)^2)/2-6(2)]

= [-9/4] - [2]

= -1/4

Area above the x-axis between roots x=1 and x=2

∫

^{2}

_{1}x^3-6x^2+11x-6 dx = (x^4)/4-2x^3+(11x^2)/2-6x |

^{2}

_{1}

= [((2)^4)/4-2(2)^3+(11(2)^2)/2-6(2)] - [((1)^4)/4-2(1)^3+(11(1)^2)/2-6(1)]

= [-2] - [-9/4]

= 1/4

= [-2] - [-9/4]

= 1/4

Xenia K.

02/02/16