What is area x=y^2-6y,x=5y-y^2 ?

The answer to area x=y^2-6y,x=5y-y^2 is 1331/24

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area x=y^2-6y,x=5y-y^2

area

x = y^{2} - 6 y, x = 5 y - y^{2}

\frac{1331}{24}

Decimal

55.45833 \dots

Solution steps

Isolate

y

for

x = y^{2} - 6 y : y = 3 + x + 9, y = 3 - x + 9

Isolate

y

for

x = 5 y - y^{2} : y = - \frac{- 5 + 25 - 4 x}{2}, y = - \frac{- 5 - 25 - 4 x}{2}

f_{1} (x) = 3 + x + 9

f_{2} (x) = 3 - x + 9

To find the intersection points solve

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

3 + x + 9 = 3 - x + 9 : x = - 9

3 + x + 9 = - \frac{- 5 + 25 - 4 x}{2} :

No Solution for

x \in R

3 + x + 9 = - \frac{- 5 - 25 - 4 x}{2} : x = - \frac{11}{4}

3 - x + 9 = - \frac{- 5 + 25 - 4 x}{2} : x = 0

3 - x + 9 = - \frac{- 5 - 25 - 4 x}{2} :

No Solution for

x \in R

- \frac{- 5 + 25 - 4 x}{2} = - \frac{- 5 - 25 - 4 x}{2} : x = \frac{25}{4}

= \int_{- 9}^{- \frac{11}{4}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{11}{4}}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{\frac{25}{4}} ∣ f_{4} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{- \frac{11}{4}} 3 + x + 9 - (3 - x + 9) d x + \int_{- \frac{11}{4}}^{0} 3 - x + 9 - (- \frac{- 5 - 25 - 4 x}{2}) d x + \int_{0}^{\frac{25}{4}} - \frac{- 5 - 25 - 4 x}{2} - (- \frac{- 5 + 25 - 4 x}{2}) d x

\int_{- 9}^{- \frac{11}{4}} 3 + x + 9 - (3 - x + 9) d x = \frac{125}{6}

\int_{- \frac{11}{4}}^{0} 3 - x + 9 - (- \frac{- 5 - 25 - 4 x}{2}) d x = \frac{331}{24}

\int_{0}^{\frac{25}{4}} - \frac{- 5 - 25 - 4 x}{2} - (- \frac{- 5 + 25 - 4 x}{2}) d x = \frac{125}{6}

= \frac{125}{6} + \frac{331}{24} + \frac{125}{6}

Simplify

= \frac{1331}{24}

What is area x=y^2-6y,x=5y-y^2 ?
The answer to area x=y^2-6y,x=5y-y^2 is 1331/24