4–x2=4–2x
x2–2x=0
x·(x–2)=0
x=0 или x–2=0 ⇒ x=2
По правилу ( см. скрин 2)
S= ∫ ^{2}_{0}( \underbrace{(4-x^2)}_{f_{2}(x)}-\underbrace{(4-2x)}_{f_{1}(x)})dx=∫ ^{2}_{0}(4-x^2-4+2x)dx=∫ ^{2}_{0}(2x-x^2)dx=
=(x^2-\frac{x^3}{3})|^{2}_{0}=(2^2-\frac{2^3}{3})-(0^2-\frac{0^3}{3})=\frac{4}{3}