What is area x=(12-y^2)/4 ,x=y ?

The answer to area x=(12-y^2)/4 ,x=y is 64/3

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area x=(12-y^2)/4 ,x=y

area

x = \frac{12 - y ^{2}}{4}, x = y

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Isolate

y

for

x = \frac{12 - y ^{2}}{4} : y = 2 - x + 3, y = - 2 - x + 3

Isolate

y

for

x = y : y = x

f_{1} (x) = 2 - x + 3

f_{2} (x) = - 2 - x + 3

To find the intersection points solve

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

2 - x + 3 = - 2 - x + 3 : x = 3

2 - x + 3 = x : x = 2

- 2 - x + 3 = x : x = - 6

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

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

\int_{- 6}^{2} - 2 - x + 3 - x d x = \frac{56}{3}

\int_{2}^{3} 2 - x + 3 - (- 2 - x + 3) d x = \frac{8}{3}

= \frac{56}{3} + \frac{8}{3}

Simplify

= \frac{64}{3}

y^{^{'}} x + 4 y x = 8, y (0) = 1

\int \frac{1}{( 1 + sin ^{2} ( x ) )} d x

x^{2} - x - 2

\frac{d}{d x} ((10 x + 2)^{x})

x y^{3}