What is area y^2=16-x,(y+2)^2=x+4 ?

The answer to area y^2=16-x,(y+2)^2=x+4 is 72

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area y^2=16-x,(y+2)^2=x+4

area

y^{2} = 16 - x, (y + 2)^{2} = x + 4

72

Solution steps

Isolate

y

for

y^{2} = 16 - x : y = 16 - x, y = - 16 - x

Isolate

y

for

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

f_{1} (x) = 16 - x

f_{2} (x) = - 16 - x

To find the intersection points solve

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

16 - x = - 16 - x : x = 16

16 - x = x + 4 - 2 : x = 12

16 - x = - x + 4 - 2 :

No Solution for

x \in R

- 16 - x = x + 4 - 2 :

No Solution for

x \in R

- 16 - x = - x + 4 - 2 : x = 0

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

= \int_{- 4}^{0} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{0}^{12} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{12}^{16} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 4}^{0} x + 4 - 2 - (- x + 4 - 2) d x + \int_{0}^{12} - 16 - x - (x + 4 - 2) d x + \int_{12}^{16} 16 - x - (- 16 - x) d x

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

\int_{0}^{12} - 16 - x - (x + 4 - 2) d x = \frac{152}{3}

\int_{12}^{16} 16 - x - (- 16 - x) d x = \frac{32}{3}

= \frac{32}{3} + \frac{152}{3} + \frac{32}{3}

Simplify

= 72

\frac{e ^{s}}{7 s + 1}

f (x) = (x e^{2 x - 4}), x = 2

n = 1 \sum \infty \frac{n - 7}{n + 7}

f (x) = sec^{5} (x)

\frac{d}{d x} (\frac{128000}{x} + x^{2})

What is area y^2=16-x,(y+2)^2=x+4 ?
The answer to area y^2=16-x,(y+2)^2=x+4 is 72