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

The answer to area y^2-4x=4,4x-y=16 is 243/8

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area y^2-4x=4,4x-y=16

area

y^{2} - 4 x = 4, 4 x - y = 16

\frac{243}{8}

Decimal

30.375

Solution steps

Isolate

y

for

y^{2} - 4 x = 4 : y = 2 1 + x, y = - 2 1 + x

Isolate

y

for

4 x - y = 16 : y = - 16 + 4 x

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

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

To find the intersection points solve

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

2 1 + x = - 2 1 + x : x = - 1

2 1 + x = - 16 + 4 x : x = \frac{21}{4}

- 2 1 + x = - 16 + 4 x : x = 3

= \int_{- 1}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{\frac{21}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 1}^{3} 2 1 + x - (- 2 1 + x) d x + \int_{3}^{\frac{21}{4}} 2 1 + x - (- 16 + 4 x) d x

\int_{- 1}^{3} 2 1 + x - (- 2 1 + x) d x = \frac{64}{3}

\int_{3}^{\frac{21}{4}} 2 1 + x - (- 16 + 4 x) d x = \frac{217}{24}

= \frac{64}{3} + \frac{217}{24}

Simplify

= \frac{243}{8}