СРОЧНО ПРОШУ!!!!! БУДУ ОЧЕНЬ БЛАГОДАРНА!!!!!!!!!!!!!!!!. Created by nasurova543. algebra-ru

СРОЧНО ПРОШУ!!!!! БУДУ ОЧЕНЬ БЛАГОДАРНА!!!!!!!!!!!!!!!!...

nasurova543 @nasurova543

July 2022 1 26 Report

СРОЧНО ПРОШУ!!!!!
БУДУ ОЧЕНЬ БЛАГОДАРНА!!!!!!!!!!!!!!!!

Answers & Comments

sangers1959

Verified answer

Y=x²+2x+2 y=6-x²
x²+2x+2=6-x²
2x²+2x-4=0 I÷2
x²+x-2=0 D=9
x₁=1 x₂=-2
S=∫₋₂¹(6-x²-x²-2x-2)dx=∫₋₂¹(4-2x-2x²)dx=(4x-x²-2x³/3) I₋₂¹=
=(4*1-1²-2*1/3)-(4*(-2)-(-2)²-2*(-2)³/3)=(4-1-2/3)-(-8-4+16/3)=
=2¹/₃-(-6²/₃)=2¹/₃+6²/₃=9.
Ответ: S=9.

y=x²-3x+4 y=4-x
x²-3x+4=4-x
x²-2x=0
x(x-2)=0
x₁=0 x₂=2
S=∫₀² (4-x-x²+3x-4)dx=∫₀² (2x-x²)dx=(x²-x³/3) I₀²=(2²-2³/3-2*0+0²)=4-2²/₃=1¹/₃,
Ответ: S=1¹/₃.

y=3/x y=3 x=3
3/x=3
3x=3
x=1
S=∫₁³ (3-3/x)dx=(3*x-3*lnx) I₁³=(3*3-3*ln3-3*1+3*ln1)=9-3*ln3-3+0=6-3*ln3.
Ответ: S=6-3*ln3≈2,7.

y=5/x x+y=6 ⇒ y=6-x
5/x=6-x
5=6x-x²
x²-6x+5=0 D=16
x₁=5 x₂=1
S=∫₁⁵ (6-x-5/x)dx=(6x-x²/2-5*lnx) I₁⁵=6*5-5²/2-5*ln5-6*1+1²/2+5*ln1=
=30-12,5-5*ln5-6+0,5+0=12-5*ln5.
Ответ: S=12-5*ln5≈3,95.

y=x²+2x+1 y=x+3
x²+2x+1=x+3
x²+x-2=0 D=9
x₁=1 x₂=-2
S=∫₋₂¹ (x+3-x²-2x-1)dx=∫₋₂¹ (2-x-x²)dx=(2x-x²/2-x³/3) I₋₂¹=
=2*1-1²/2-1³/3-2(-2)-(-2)²/2+(-2)³/3=2-1/2-1/3+4+2-8/3=8-3,5=4,5.
Ответ: S=4,5.
y=4-x² y=2-x
4-x²=2-x
x²-x-2=0 D=9
x₁=2 x₂=-1
S=∫₋₁² (4-x²-2+x)dx=∫₋₁² (2+x-x²)dx=(2x+x²/2-x³/3) I₋₁²=
=2*2+2²/2-2³/3-2(-1)-(-1)²/2+(-1)³/3=4+2-2²/₃+2-1/2-1/3=8-3¹/₂=4¹/₂=4,5.
Ответ: S=4,5.

y=x² y=2x-x²
x²=2x-x²
2x²-2x=0 I÷2
x(x-1)=0
x₁=0 x₂=1
S=∫₀¹ (2x-x²-x²)dx=∫₀¹ (2x-2x²)dx=(x²-2x³/3) I₀¹=
=1²-2*1³/3-0²+2*0³/3=1-2/3=1/3.
Ответ: S=1/3.

y=6-x² y=5
6-x²=5
x²=1
x₁=1 x₂=-1
S=∫₋₁¹(6-x²-5)dx=∫₋₁¹(1-x²)dx=(x-x³/3) I₋₁¹=1-1³/3-(-1)+(-1)³/3=2*2/3=1¹/₃.
Ответ: S=1¹/₃.

2*sin²(3α)+5*sin(3π/2-α)+2*cos²(3α)=2(sin²(3α)+cos²(3α))+5*(-cosα)=
=2*1-5*cosα=2-5*0,2=2-1=1.

2 votes Thanks 1

sangers1959 sin(α+π/3)-(√3/2)*cosα=sinα*cos(π/3)+sin(π/3)*cosα-(√3/2)*cosα=
=(1/2)*sinα+(√3/2)*cosα-(√3/2)*cosα=sinα/2.
Ответ: sinα/2.

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