What is area y=x^3,x=y^2-2 ?

The answer to area y=x^3,x=y^2-2 is 4/3+2.88161…

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area y=x^3,x=y^2-2

area

y = x^{3}, x = y^{2} - 2

\frac{4}{3} + 2.88161 \dots

Decimal

4.21494 \dots

Solution steps

Isolate

y

for

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

f_{1} (x) = x^{3}

f_{2} (x) = x + 2

To find the intersection points solve

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

x^{3} = x + 2 : x \approx 1.21486 \dots

x^{3} = - x + 2 : x = - 1

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

= \int_{- 2}^{- 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{1.21486 \dots} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{- 1} x + 2 - (- x + 2) d x + \int_{- 1}^{1.21486 \dots} x^{3} - x + 2 d x

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

\int_{- 1}^{1.21486 \dots} x^{3} - x + 2 d x = 2.88161 \dots

= \frac{4}{3} + 2.88161 \dots

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 5)

\int \frac{7}{( x - 1 ) x ^{2} - 2 x - 48} d x

f (x) = \frac{1}{x ^{4}}

\int arccot (12 y) d y

x \to - 5 lim ((4 - x)^{\frac{3}{2}})