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

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

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

area

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

\frac{243}{8}

Solution steps

Isolate

y

for

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

f_{1} (x) = 4 x - 16

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

To find the intersection points solve

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

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

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

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

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

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

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

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

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

Simplify

= \frac{243}{8}

y = arctan (7 x^{2} - 1)

\int x (cos^{2} (x)) d x

y = x^{2} - 2 x - 8

\int (x^{2} + y) d x

\frac{d}{d y} (\frac{1}{y ^{2} + 1})