Question

Tentukan luas bidang yang dibatasi oleh kurva x=5y^2 dan garis x=1+y^2

Answer 1

5Y^2 = 1 + Y^2

4Y^2 = 1

Y = 1/2 dan -1/2

L = integral dr ( 1 + Y^2 -(5Y^2)) dg batas 1/2 dan - 1/2

jd L = [ Y - 4/3 Y^3]
L = (1/2 - 4/3(1/2)^3) - (-1/2 - 4/3(-1/2)^3)

L = 2/6 - (-2/6) = 4/6

Answer 2

5Y^2 = 1 + Y^2

4Y^2 = 1

Y = 1/2 dan -1/2

L = integral dr ( 1 + Y^2 -(5Y^2)) dg batas 1/2 dan - 1/2

jd L = [ Y - 4/3 Y^3] 
L = (1/2 - 4/3(1/2)^3) - (-1/2 - 4/3(-1/2)^3)

L = 2/6 - (-2/6) = 4/6