Calculate the area of the shape bounded by the lines y = x ^ 2-4x + 5 and y = 5.

Let’s find the points of intersection of the lines analytically:
x ^ 2 – 4x + 5 = 5,
x (x – 4) = 0,
x = 0, x = 4
Lower limit of integration: x = 0
Upper limit of integration: x = 4
The desired figure is bounded from above by the straight line y = 5, and from below by the parabola y = x ^ 2-4x + 5, then:
(5 – x ^ 2 + 4x + 5) dx = (- x ^ 2 + 4x) dx = -x ^ 3/3 + 2x ^ 2
Since the boundaries are equal to x = 0, x = 4, respectively, we substitute only the upper limit, because the lower one gives zero, into the resulting expression:
-64/3 + 32 = 32/3
Answer: the area of the figure is 32/3