Calculus question

Question

The area between the two curves y = x²-2x+4 and 2y=x-2 bound by x = 0 and x =3 is what?

 

Help!?

Answer 1

If we graph the functions y = x² - 2x + 4 and y = (1/2)x - 1 on the interval [0,3], we see that the graph of the first equation always lies above the graph of the second equation.  

 

Area of region = ∫_{(from 0 to 3)}[upper boundary - lower boundary] dx

 

                     = ∫_{(from 0 to 3)}[(x²-2x+4) - ((1/2)x - 1)] dx

 

                     = ∫_{(from 0 to 3)}[x² - (5/2x) + 5]dx

 

                     = [(1/3)x³ - (5/4)x² + 5x]_{(from 0 to 3)}

 

                     = 9 - 45/4 + 15

 

                     = 24 - 45/4 = 51/4 = 12.75

Answer 2

They do not interest.

 

Per the second function y = (x-2)/2

 

Setting them equal:

    x^2 - 2x +4 = (x-2)/2

 

 2x^2 - 4x + 8 = x-2

 

2x^2 - 5x + 6 = 0

 

The discrimminant B^2 - 4AC = 25 - 4(2)(6) = 25 - 48 = -23

 

So no real roots.