Question

Hitunglah luas yang dibatasi oleh kurva y=√(x+2), sumbu x, garis x=2 y≥0

Answer 1

Verified answer

Y = √(x + 2)

Batas integral
y = 0 ---> x = -2

Luas
= ∫√(x + 2) dx [2...-2]
= ∫ (x + 2)^1/2 d(x + 2)
= (1/(1/2 + 1))(x + 2)^3/2
= (2/3)(x + 2)^3/2
= (2/3)(x + 2)^3/2
= (2/3)((2 + 2)^3/2 - (-2 + 2)^3/2)
= (2/3)(2^3 - 0)
= 16/3 satuan
= 5,33 satuan luas

Answer 2

Pertama, tentukan titik titik kurvanya
y = √(x+2)
y = 0, x = -2
y = 2, x = 2
y = -2, x = 2

trus gambar kurvanya *ada di foto

daerah yg diarsir itu luasnya. Batas atas 2 dan batas bawah 0

Batas garis atasnya y=√(x+2); garis bawah x

₀∫² (√(x+2) - (x)) dx

=(2/3) (x + 2)^(3/2) - (1/2) (x^2)]²₀
=[(2/3)(2+2)^(3/2) - (2/3)(0+2)^(3/2)]-[(2^2 /2)-0)]
=[(2/3) (8) - (2/3) √8] - [2]
=((16 - √8) / 3) - 2
=((16 - √8) / 3) - (6/3)
= (10 - √8) / 3 satuan

semoga membantu :D