1) (х·у+3·х2)·d·х+(х–у)d·у
А(1;2), В (3;4)
2) ( х2+у2)d·х + (2·х–2·у)·d·y
А(1;3), В (2;4)
3) (6·х·у2+3·у2)d·х+(6·х·у+4·х3)·d·y
А(–3;2), В (–1;–3)
(x–1)/(3–1)=(y–2)(4–2)
y=x+1
dy=dx
∫ (х·у+3х2)·d·х+(х–у)d·у= ∫31 (x·(x+1)+3x2)dx+(x–x–1)dx)=
=∫31 (x2+x+3x2–1)dx=∫31 (4x2+x–1)dx=
=((4x3/3)+(x2/2)–x)|31=
=(4/3)·(33–13)+(1/2)·(32–12)–(3–1)=
=(4/3)·26+4–2= 110/3