Calculate the area of the shape bounded by the lines y = x ^ 2-4x + 6 and y = 6.

1. Let’s find the difference of two functions:

f (x) = x ^ 2 – 4x + 6;
g (x) = 6;
h (x) = g (x) – f (x) = 6 – x ^ 2 + 4x – 6 = 4x – x ^ 2 = x (4 – x).
2. Zeros of function:

x (4 – x) = 0;
[x = 0;
[4 – x = 0;
[x = 0;
[x = 4.
3. The area of the figure bounded by the graphs of the functions f (x) and g (x) is equal to a definite integral of the function h (x) in the range from 0 to 4:

F (x) = ∫h (x) dx;
F (x) = ∫ (4x – x ^ 2) dx = 2x ^ 2 – x ^ 3/3;
F (0) = 2 * 0 ^ 2 – 0 ^ 3/3 = 0;
F (4) = 2 * 4 ^ 2 – 4 ^ 3/3 = 32 – 64/3 = (96 – 64) / 3 = 32/3;
S = F (4) – F (0) = 32/3 – 0 = 10 2/3.
Answer: 10 2/3.