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

The answer to area x=y^2-4y,x=y-y^2 is 125/24

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

area

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

\frac{125}{24}

Decimal

5.20833 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x = y - y^{2} : y = - \frac{- 1 + 1 - 4 x}{2}, y = - \frac{- 1 - 1 - 4 x}{2}

f_{1} (x) = 2 + x + 4

f_{2} (x) = 2 - x + 4

To find the intersection points solve

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

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

2 + x + 4 = - \frac{- 1 + 1 - 4 x}{2} :

No Solution for

x \in R

2 + x + 4 = - \frac{- 1 - 1 - 4 x}{2} : x = - \frac{15}{4}

2 - x + 4 = - \frac{- 1 + 1 - 4 x}{2} : x = 0

2 - x + 4 = - \frac{- 1 - 1 - 4 x}{2} :

No Solution for

x \in R

- \frac{- 1 + 1 - 4 x}{2} = - \frac{- 1 - 1 - 4 x}{2} : x = \frac{1}{4}

= \int_{- 4}^{- \frac{15}{4}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{15}{4}}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{\frac{1}{4}} ∣ f_{4} (x) - f_{3} (x) ∣ d x

= \int_{- 4}^{- \frac{15}{4}} 2 + x + 4 - (2 - x + 4) d x + \int_{- \frac{15}{4}}^{0} 2 - x + 4 - (- \frac{- 1 - 1 - 4 x}{2}) d x + \int_{0}^{\frac{1}{4}} - \frac{- 1 - 1 - 4 x}{2} - (- \frac{- 1 + 1 - 4 x}{2}) d x

\int_{- 4}^{- \frac{15}{4}} 2 + x + 4 - (2 - x + 4) d x = \frac{1}{6}

\int_{- \frac{15}{4}}^{0} 2 - x + 4 - (- \frac{- 1 - 1 - 4 x}{2}) d x = \frac{39}{8}

\int_{0}^{\frac{1}{4}} - \frac{- 1 - 1 - 4 x}{2} - (- \frac{- 1 + 1 - 4 x}{2}) d x = \frac{1}{6}

= \frac{1}{6} + \frac{39}{8} + \frac{1}{6}

Simplify

= \frac{125}{24}

2 y^{^{''}} - 15 y^{^{'}} - 8 y = 0

\int_{- 1}^{1} x - 1 - x^{2} d x

y = (x^{6} - 10) x

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n = 3 \sum \infty \frac{1}{n ^{2} - 1}