Find the area of the figure bounded by the parabola y = x ^ 2-6x + 8 and the Ox axis.

we obtain the system of equations
y = x ^ 2-6x + 8
y = 0
we will solve it
and find the intersection points of the parabola and the line
x squared – 6 * x + 8 = 0
Find the discriminant of the quadratic equation:
D = b squared – 4ac = (-6) squared – 4 1 8 = 36 – 32 = 4
Since the discriminant is greater than zero, the quadratic equation has two real roots:
x1 = (6 – √4) / (2 1) = (6 – 2) / 2 = 4/2 = 2
x2 = (6 + √4) / (2 1) = (6 + 2) 2 = 8/2 = 4

intersection points (2; 0) and (4; 0)
integral = x cubed / 3 – 6 * x squared / 2 + 8 x = x cubed / 3 – 3 * x squared + 8 * x

find a definite integral from 2 to 4, this will be the area of ​​the figure bounded by a straight line and a parabola
definite integral from 2 to 4 ooo = 4 * 4 * 4/3 – 3 * 4 * 4 + 8 * 4 – (2 * 2 * 2/3 – 3 * 2 * 2 + 8 * 2) = 64/3 – 64 +32 – 8/3 + 12 – 16 = 56/3 – 36 = (56 – 3 * 36) / 3 = – 52/3
area is = 52/3