What is area x=y^2-1,x=y ?

The answer to area x=y^2-1,x=y is (sqrt(2)(3-sqrt(5))^{3/2}+sqrt(2)(3+sqrt(5))^{3/2}-3sqrt(5))/6

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area x=y^2-1,x=y

area

x = y^{2} - 1, x = y

\frac{2 ( 3 - 5 ) ^{\frac{3}{2}} + 2 ( 3 + 5 ) ^{\frac{3}{2}} - 3 5}{6}

Decimal Notation

1.86338 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x = y : y = x

f_{1} (x) = x + 1

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

To find the intersection points solve

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

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

x + 1 = x : x = \frac{1 + 5}{2}

- x + 1 = x : x = \frac{1 - 5}{2}

= \int_{- 1}^{\frac{1 - 5}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{1 - 5}{2}}^{\frac{1 + 5}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 1}^{\frac{1 - 5}{2}} x + 1 - (- x + 1) d x + \int_{\frac{1 - 5}{2}}^{\frac{1 + 5}{2}} x + 1 - x d x

\int_{- 1}^{\frac{1 - 5}{2}} x + 1 - (- x + 1) d x = \frac{2 ( 3 - 5 ) ^{\frac{3}{2}}}{3}

\int_{\frac{1 - 5}{2}}^{\frac{1 + 5}{2}} x + 1 - x d x = \frac{( 3 + 5 ) ^{\frac{3}{2}} - ( 3 - 5 ) ^{\frac{3}{2}}}{3 2} - \frac{5}{2}

= \frac{2 ( 3 - 5 ) ^{\frac{3}{2}}}{3} + \frac{( 3 + 5 ) ^{\frac{3}{2}} - ( 3 - 5 ) ^{\frac{3}{2}}}{3 2} - \frac{5}{2}

Simplify

= \frac{2 ( 3 - 5 ) ^{\frac{3}{2}} + 2 ( 3 + 5 ) ^{\frac{3}{2}} - 3 5}{6}

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x + 1)

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

\int tan^{2} (8 x) d x

y = \frac{e ^{x}}{x - 1}

\frac{d y}{d x} = \frac{( e ^{- y} + e ^{- 5 x - y} )}{e ^{x} y}

What is area x=y^2-1,x=y ?
The answer to area x=y^2-1,x=y is (sqrt(2)(3-sqrt(5))^{3/2}+sqrt(2)(3+sqrt(5))^{3/2}-3sqrt(5))/6