f(x) = x^3 - 4·x^2 + 6·x
t(x) = f'(1) * (x - 1) + f(1) = x + 2
d(x) = f(x) - t(x) = (x^3 - 4·x^2 + 6·x) - (x + 2) = x^3 - 4·x^2 + 5·x - 2
D(x) = x^4/4 - 4·x^3/3 + 5·x^2/2 - 2·x
Schnittpunkte d(x) = 0
x^3 - 4·x^2 + 5·x - 2 = 0
x = 2 ∨ x = 1
D(2) - D(1) = - 2/3 - (- 7/12) = -1/12
Die Fläche beträgt 1/12