What is area x= 3/y ,x= y/9 ,x=9y ?

The answer to area x= 3/y ,x= y/9 ,x=9y is 12ln(3)

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area x= 3/y ,x= y/9 ,x=9y

area

x = \frac{3}{y}, x = \frac{y}{9}, x = 9 y

12 ln (3)

Decimal

13.18334 \dots

Solution steps

Isolate

y

for

x = \frac{3}{y} : y = \frac{3}{x}; x \neq = 0

Isolate

y

for

x = \frac{y}{9} : y = 9 x

Isolate

y

for

x = 9 y : y = \frac{x}{9}

f_{1} (x) = \frac{3}{x}

f_{2} (x) = 9 x

To find the intersection points solve

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

\frac{3}{x} = 9 x : x = \frac{1}{3}, x = - \frac{1}{3}

\frac{3}{x} = \frac{x}{9} : x = 33, x = - 33

9 x = \frac{x}{9} : x = 0

= \int_{- 33}^{- \frac{1}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{3}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{1}{3}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3}}^{33} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 33}^{- \frac{1}{3}} \frac{3}{x} - \frac{x}{9} d x + \int_{- \frac{1}{3}}^{0} \frac{x}{9} - 9 x d x + \int_{0}^{\frac{1}{3}} 9 x - \frac{x}{9} d x + \int_{\frac{1}{3}}^{33} \frac{3}{x} - \frac{x}{9} d x

\int_{- 33}^{- \frac{1}{3}} \frac{3}{x} - \frac{x}{9} d x = 6 ln (3) - \frac{40}{27}

\int_{- \frac{1}{3}}^{0} \frac{x}{9} - 9 x d x = \frac{40}{27}

\int_{0}^{\frac{1}{3}} 9 x - \frac{x}{9} d x = \frac{40}{27}

\int_{\frac{1}{3}}^{33} \frac{3}{x} - \frac{x}{9} d x = 6 ln (3) - \frac{40}{27}

= 6 ln (3) - \frac{40}{27} + \frac{40}{27} + \frac{40}{27} + 6 ln (3) - \frac{40}{27}

Simplify

= 12 ln (3)

\int 20

\frac{d}{d x} (sin (x + 3))

f (x) = ln (x - 2)

y^{^{'}} - \frac{2}{x} y = \frac{y ^{3}}{x ^{3}}

h \to 0 lim (\frac{( e ^{2 h} - 1 )}{h})

What is area x= 3/y ,x= y/9 ,x=9y ?
The answer to area x= 3/y ,x= y/9 ,x=9y is 12ln(3)