What is area x=2y-y^2,x=y^2-4y ?

The answer to area x=2y-y^2,x=y^2-4y is 9

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area x=2y-y^2,x=y^2-4y

area

x = 2 y - y^{2}, x = y^{2} - 4 y

9

Solution steps

Isolate

y

for

x = 2 y - y^{2} : y = 1 - - x + 1, y = - x + 1 + 1

Isolate

y

for

x = y^{2} - 4 y : y = 2 + x + 4, y = 2 - x + 4

f_{1} (x) = 1 - - x + 1

f_{2} (x) = - x + 1 + 1

To find the intersection points solve

f_{i} (x) = f_{j} (x)

1 - - x + 1 = - x + 1 + 1 : x = 1

1 - - x + 1 = 2 + x + 4 :

No Solution for

x \in R

1 - - x + 1 = 2 - x + 4 : x = 0

- x + 1 + 1 = 2 + x + 4 : x = - 3

- x + 1 + 1 = 2 - x + 4 :

No Solution for

x \in R

2 + x + 4 = 2 - x + 4 : x = - 4

= \int_{- 4}^{- 3} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{- 3}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 4}^{- 3} 2 + x + 4 - (2 - x + 4) d x + \int_{- 3}^{0} - x + 1 + 1 - (2 - x + 4) d x + \int_{0}^{1} 1 - - x + 1 - (- x + 1 + 1) d x

\int_{- 4}^{- 3} 2 + x + 4 - (2 - x + 4) d x = \frac{4}{3}

\int_{- 3}^{0} - x + 1 + 1 - (2 - x + 4) d x = \frac{19}{3}

\int_{0}^{1} 1 - - x + 1 - (- x + 1 + 1) d x = \frac{4}{3}

= \frac{4}{3} + \frac{19}{3} + \frac{4}{3}

Simplify

= 9

x^{2} - x, x = 4

\frac{d}{d x} (2 cos (2 x) - 1)

x \to 4 lim (x - x^{2})

\frac{d}{d y} (4 arctan (\frac{x}{y}))

\frac{\partial}{\partial x} (4 y^{2} + 7 x^{2})