What is area x=4y-y^2,x=y^2-7y ?

The answer to area x=4y-y^2,x=y^2-7y is 1331/24

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area x=4y-y^2,x=y^2-7y

area

x = 4 y - y^{2}, x = y^{2} - 7 y

\frac{1331}{24}

Decimal

55.45833 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x = y^{2} - 7 y : y = \frac{7 + 49 + 4 x}{2}, y = \frac{7 - 49 + 4 x}{2}

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

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

To find the intersection points solve

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

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

2 - - x + 4 = \frac{7 + 49 + 4 x}{2} :

No Solution for

x \in R

2 - - x + 4 = \frac{7 - 49 + 4 x}{2} : x = 0

- x + 4 + 2 = \frac{7 + 49 + 4 x}{2} : x = - \frac{33}{4}

- x + 4 + 2 = \frac{7 - 49 + 4 x}{2} :

No Solution for

x \in R

\frac{7 + 49 + 4 x}{2} = \frac{7 - 49 + 4 x}{2} : x = - \frac{49}{4}

= \int_{- \frac{49}{4}}^{- \frac{33}{4}} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{- \frac{33}{4}}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- \frac{49}{4}}^{- \frac{33}{4}} \frac{7 + 49 + 4 x}{2} - \frac{7 - 49 + 4 x}{2} d x + \int_{- \frac{33}{4}}^{0} - x + 4 + 2 - \frac{7 - 49 + 4 x}{2} d x + \int_{0}^{4} 2 - - x + 4 - (- x + 4 + 2) d x

\int_{- \frac{49}{4}}^{- \frac{33}{4}} \frac{7 + 49 + 4 x}{2} - \frac{7 - 49 + 4 x}{2} d x = \frac{32}{3}

\int_{- \frac{33}{4}}^{0} - x + 4 + 2 - \frac{7 - 49 + 4 x}{2} d x = \frac{273}{8}

\int_{0}^{4} 2 - - x + 4 - (- x + 4 + 2) d x = \frac{32}{3}

= \frac{32}{3} + \frac{273}{8} + \frac{32}{3}

Simplify

= \frac{1331}{24}

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What is area x=4y-y^2,x=y^2-7y ?
The answer to area x=4y-y^2,x=y^2-7y is 1331/24