What is area x+y=-3,x+9=y^2 ?

The answer to area x+y=-3,x+9=y^2 is 125/6

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area x+y=-3,x+9=y^2

area

x + y = - 3, x + 9 = y^{2}

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x + 9 = y^{2} : y = x + 9, y = - x + 9

f_{1} (x) = - 3 - x

f_{2} (x) = x + 9

To find the intersection points solve

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

- 3 - x = x + 9 : x = - 5

- 3 - x = - x + 9 : x = 0

x + 9 = - x + 9 : x = - 9

= \int_{- 9}^{- 5} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 5}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{- 5} x + 9 - (- x + 9) d x + \int_{- 5}^{0} - 3 - x - (- x + 9) d x

\int_{- 9}^{- 5} x + 9 - (- x + 9) d x = \frac{32}{3}

\int_{- 5}^{0} - 3 - x - (- x + 9) d x = \frac{61}{6}

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

Simplify

= \frac{125}{6}

\frac{d}{d x} (cos (x - \frac{π}{2}))

\frac{e ^{5 x}}{x ^{20}}

y = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

- 6 x (sin (x) + cos (x))

sin (2 x) d x + cos (8 y) d y = 0, y (π^{2}) = π^{8}