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Popular >

area x=y-y^2,x=y^2-3

Solution

area

x = y - y^{2}, x = y^{2} - 3

Solution

\frac{125}{24}

+1

Solution steps

Isolate

y

for

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

Isolate

y

for

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

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

f_{2} (x) = - \frac{- 1 - 1 - 4 x}{2}

To find the intersection points solve

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

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

- \frac{- 1 + 1 - 4 x}{2} = x + 3 :

No Solution for

x \in R

- \frac{- 1 + 1 - 4 x}{2} = - x + 3 : x = - 2

- \frac{- 1 - 1 - 4 x}{2} = x + 3 : x = - \frac{3}{4}

- \frac{- 1 - 1 - 4 x}{2} = - x + 3 :

No Solution for

x \in R

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

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

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

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

\int_{- 2}^{- \frac{3}{4}} - \frac{- 1 + 1 - 4 x}{2} - x + 3 d x = \frac{61}{24}

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

= \frac{4}{3} + \frac{61}{24} + \frac{4}{3}

Simplify

= \frac{125}{24}

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Frequently Asked Questions (FAQ)

What is area x=y-y^2,x=y^2-3 ?
The answer to area x=y-y^2,x=y^2-3 is 125/24