What is area x=10y-y^2,x=y^2-4y ?

The answer to area x=10y-y^2,x=y^2-4y is 343/3

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area x=10y-y^2,x=y^2-4y

area

x = 10 y - y^{2}, x = y^{2} - 4 y

\frac{343}{3}

Decimal Notation

114.33333 \dots

Solution steps

Isolate

y

for

x = 10 y - y^{2} : y = 5 - - x + 25, y = - x + 25 + 5

Isolate

y

for

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

f_{1} (x) = 5 - - x + 25

f_{2} (x) = - x + 25 + 5

To find the intersection points solve

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

5 - - x + 25 = - x + 25 + 5 : x = 25

5 - - x + 25 = 2 + x + 4 :

No Solution for

x \in R

5 - - x + 25 = 2 - x + 4 : x = 0

- x + 25 + 5 = 2 + x + 4 : x = 21

- x + 25 + 5 = 2 - x + 4 :

No Solution for

x \in R

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

= \int_{- 4}^{0} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{0}^{21} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{21}^{25} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 4}^{0} 2 + x + 4 - (2 - x + 4) d x + \int_{0}^{21} 5 - - x + 25 - (2 + x + 4) d x + \int_{21}^{25} - x + 25 + 5 - (5 - - x + 25) d x

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

\int_{0}^{21} 5 - - x + 25 - (2 + x + 4) d x = 93

\int_{21}^{25} - x + 25 + 5 - (5 - - x + 25) d x = \frac{32}{3}

= \frac{32}{3} + 93 + \frac{32}{3}

Simplify

= \frac{343}{3}

\int \frac{- 1000}{x ^{2}} d x

\frac{s ^{2} + s}{s ^{2} + s + 1}

y^{^{''}} - 3 y^{^{'}} + 7 y = x e^{x}

\int \frac{8}{12 - x ^{2} - 4 x} d x

3 x^{2}

What is area x=10y-y^2,x=y^2-4y ?
The answer to area x=10y-y^2,x=y^2-4y is 343/3