What is area y=x,4x+y^2=12 ?

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

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area y=x,4x+y^2=12

area

y = x, 4 x + y^{2} = 12

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Isolate

y

for

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

f_{1} (x) = x

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

To find the intersection points solve

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

x = 2 3 - x : x = 2

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

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

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

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

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

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

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

Simplify

= \frac{64}{3}

{\frac{5}{s ^{2} + 49}}

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