Question

the area of the region enclosed by the graphs of y =x and y = x^2-3x+3 is?
a) 2/3
b) 1
c) 4/3
d) 2
e) 14/3

Answer 1

first write the equation of the parábola in vertex form

Answer 2

ah should be y = (x-1)^2 + 1?

Answer 3

i dont see why to put it in vertex form

Answer 4

find intercept points:
x^2-3x+3-x=0
x=1,x=3
\[\int\limits_{1}^{3}(x- x^2+3x-3)dx=4/3\] the answer c

Answer 5

thanks! i got that far just mixed things up