What is area x=(y^2-1)/3 ,x=3+y ?

The answer to area x=(y^2-1)/3 ,x=3+y is 343/18

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area x=(y^2-1)/3 ,x=3+y

area

x = \frac{y ^{2} - 1}{3}, x = 3 + y

\frac{343}{18}

Decimal Notation

19.05555 \dots

Solution steps

Isolate

y

for

x = \frac{y ^{2} - 1}{3} : y = 3 x + 1, y = - 3 x + 1

Isolate

y

for

x = 3 + y : y = x - 3

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

f_{2} (x) = - 3 x + 1

To find the intersection points solve

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

3 x + 1 = - 3 x + 1 : x = - \frac{1}{3}

3 x + 1 = x - 3 : x = 8

- 3 x + 1 = x - 3 : x = 1

= \int_{- \frac{1}{3}}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{8} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{1}{3}}^{1} 3 x + 1 - (- 3 x + 1) d x + \int_{1}^{8} 3 x + 1 - (x - 3) d x

\int_{- \frac{1}{3}}^{1} 3 x + 1 - (- 3 x + 1) d x = \frac{32}{9}

\int_{1}^{8} 3 x + 1 - (x - 3) d x = \frac{31}{2}

= \frac{32}{9} + \frac{31}{2}

Simplify

= \frac{343}{18}

\frac{d}{d x} (- 2 ln (x) + x^{2} - 4 + e^{x} + x^{e})

\frac{\partial}{\partial x} (2 yz)

x \to 6 - lim (e^{\frac{3}{6 - x}})

\frac{d y}{d x} = y x

\frac{d}{d x} ((10 - 5 x^{2})^{\frac{4}{3}})

What is area x=(y^2-1)/3 ,x=3+y ?
The answer to area x=(y^2-1)/3 ,x=3+y is 343/18