Find the area bounded by y=x^2-4x+3 and y=x+4 to the newrest integer.
need their intersection:
x^2 - 4x + 3 = x + 4
x^2 - 5x - 1 = 0
was hoping for "nicer" values, but anyway....
x = (5 ± √29)/2
height of our region within those x values
= x+4 - (x^2 - 4x + 3)
= 1 + 5x - x^2
area = ∫(1 + 5x - x^2)dx from x = (5-√29)/2 to (5+√29)/2
= [x + (5/2)x^2 - (1/3)x^3 ] for x = (the above)
use your calculator to do the arithmetic