ответ: S=12 кв. ед.
Объяснение:
y=(22/3)-(x+1)² y=x/3 x₁=-1 x₂=1.
S=₋₁∫¹((22/3)-(x+1)²-(x/3))dx=₋₁∫¹((22/3)-x²-2x-1-(x/3))dx=
₋₁∫¹((19/3)-x²-(7x/3))dx=(19x/3)-(x³/3)-(7x²/6) ₋₁|¹=
=(19*1/3)-(1³/3)-(7*1²/6)-((19*(-1)/3)-(-1)³/3)-(7*(-1)²/6))=
=(19/3)-(1/3)-(7/6)-((-19/3)+(1/3)-(7/6))=6-(7/6)-(-6-(7/6))=
=6-(7/6)+6+(7/6)=6+6=12.
ответ: S=12 кв. ед.
Объяснение:
y=(22/3)-(x+1)² y=x/3 x₁=-1 x₂=1.
S=₋₁∫¹((22/3)-(x+1)²-(x/3))dx=₋₁∫¹((22/3)-x²-2x-1-(x/3))dx=
₋₁∫¹((19/3)-x²-(7x/3))dx=(19x/3)-(x³/3)-(7x²/6) ₋₁|¹=
=(19*1/3)-(1³/3)-(7*1²/6)-((19*(-1)/3)-(-1)³/3)-(7*(-1)²/6))=
=(19/3)-(1/3)-(7/6)-((-19/3)+(1/3)-(7/6))=6-(7/6)-(-6-(7/6))=
=6-(7/6)+6+(7/6)=6+6=12.