Определенный Интеграл. y=1–x2 x=0 x=√y–2 x=1
x=√y–2 x2=y–2 y=x2+2 = ∫01(x2+2–(1–x2))dx=∫01(x2+2–1+x2)dx=∫01(2x2+1)dx=((x3/3)+x)|01= =(1/3)+1=(4/3)