What is area x+y^2=20,x+y=0 ?

The answer to area x+y^2=20,x+y=0 is 243/2

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area x+y^2=20,x+y=0

area

x + y^{2} = 20, x + y = 0

\frac{243}{2}

Decimal Notation

121.5

Solution steps

Isolate

y

for

x + y^{2} = 20 : y = 20 - x, y = - 20 - x

Isolate

y

for

x + y = 0 : y = - x

f_{1} (x) = 20 - x

f_{2} (x) = - 20 - x

To find the intersection points solve

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

20 - x = - 20 - x : x = 20

20 - x = - x : x = - 5

- 20 - x = - x : x = 4

= \int_{- 5}^{4} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{4}^{20} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 5}^{4} 20 - x - (- x) d x + \int_{4}^{20} - 20 - x - 20 - x d x

\int_{- 5}^{4} 20 - x - (- x) d x = \frac{217}{6}

\int_{4}^{20} - 20 - x - 20 - x d x = \frac{256}{3}

= \frac{217}{6} + \frac{256}{3}

Simplify

= \frac{243}{2}

f (x) = x^{2} - 3 x + 4

x \to - 8 lim (4 x - 2)

\frac{\partial}{\partial y} (\frac{x}{( y + z )})

\frac{\partial}{\partial x} (x ln (x))

\frac{d}{d x} (2 x + \frac{200}{x})