Find the area between x= y^2- 4y and x = y-4

Find the area between x= y^2- 4y and x = y-4

Thanks for the help!

Set each equation equal to each other to find points of intersection: y^2 - 4y = y - 4

Subtract y to each side: y^2 - 5y = -4

Add 4 to each side: y^2 - 5y + 4 = 0

Factor: (y - 4) * (y - 1) = 0

Set each factor equal to 0 and solve for y separately.

y - 4 = 0
y - 1 = 0

Add 4 to each side: y = 4
Add 1 to each side: y = 1

Area = Integral from 1 to 4 (((y - 4) - (y^2 - 4y)) dy)

Simplify: Area = Integral from 1 to 4 ((-y^2 + 5y - 4) dy)

Integrate: Area = -(1/3)y^3 + (5/2)y^2 - 4y from 1 to 4

Substitute 4 for y and 1 for y then take the difference: Area = -(1/3)(4)^3 + (5/2)(4)^2 - 4(4) + (1/3)(1)^3 - (5/2)(1)^2 + 4(1)

Simplify: Area = -(1/3)(64) + (5/2)(16) - 16 + (1/3) - (5/2) + 4

Simplify: Area = -(64/3) + 40 - 16 + (1/3) - (5/2) + 4

Simplify: Area = -(63/3) - (5/2) + 28

Simplify: Area = -21 - (5/2) + 28

Simplify: Area = 7 - (5/2)

Common denominators: Area = (14 - 5) / 2

Simplify: Area = 9 / 2

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