Cubic curve integral - area enclosing curve and x-axis

Question

The curve is x^3-6x^2+11x-6. I have to prove that the two regions enclosed between the x-axis and the curve are equal. Given is, that the curve cuts the x-axis at x=1,x=2 and x=3. I tried finding the area by doing the integral between 1 and 2, and then 2 and 3, but I always get the wrong answer. It should be 1/4 square unit. Please help me with this? 

Answer 1

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

 

                                                                                

∫³₂    x^3-6x^2+11x-6 dx = (x^4)/4-2x^3+(11x^2)/2-6x |³₂

 

=  [((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

 

∫²₁ x^3-6x^2+11x-6 dx = (x^4)/4-2x^3+(11x^2)/2-6x |²₁

 

= [((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