What is area x+y^2=6,x+y=0 ?

The answer to area x+y^2=6,x+y=0 is 125/6

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area x+y^2=6,x+y=0

area

x + y^{2} = 6, x + y = 0

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x + y = 0 : y = - x

f_{1} (x) = 6 - x

f_{2} (x) = - 6 - x

To find the intersection points solve

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

6 - x = - 6 - x : x = 6

6 - x = - x : x = - 3

- 6 - x = - x : x = 2

= \int_{- 3}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{2}^{6} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 3}^{2} 6 - x - (- x) d x + \int_{2}^{6} - 6 - x - 6 - x d x

\int_{- 3}^{2} 6 - x - (- x) d x = \frac{61}{6}

\int_{2}^{6} - 6 - x - 6 - x d x = \frac{32}{3}

= \frac{61}{6} + \frac{32}{3}

Simplify

= \frac{125}{6}

y = (x^{6} - 7 x + 4) (x^{4} - 5 x)

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

(\frac{y}{x} + 3 x) d x + (ln (x) - 2) d y = 0

\int \frac{x ^{2} - 2 x}{x ^{3} - 3 x ^{2}} d x

- 3 t (6 t^{3} - 1)^{8}