What is area x= 4/y ,x= y/8 ,x=8y ?

The answer to area x= 4/y ,x= y/8 ,x=8y is 24ln(2)

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area x= 4/y ,x= y/8 ,x=8y

area

x = \frac{4}{y}, x = \frac{y}{8}, x = 8 y

24 ln (2)

Decimal Notation

16.63553 \dots

Solution steps

Isolate

y

for

x = \frac{4}{y} : y = \frac{4}{x}; x \neq = 0

Isolate

y

for

x = \frac{y}{8} : y = 8 x

Isolate

y

for

x = 8 y : y = \frac{x}{8}

f_{1} (x) = \frac{4}{x}

f_{2} (x) = 8 x

To find the intersection points solve

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

\frac{4}{x} = 8 x : x = \frac{1}{2}, x = - \frac{1}{2}

\frac{4}{x} = \frac{x}{8} : x = 42, x = - 42

8 x = \frac{x}{8} : x = 0

= \int_{- 42}^{- \frac{1}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{2}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{42} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 42}^{- \frac{1}{2}} \frac{4}{x} - \frac{x}{8} d x + \int_{- \frac{1}{2}}^{0} \frac{x}{8} - 8 x d x + \int_{0}^{\frac{1}{2}} 8 x - \frac{x}{8} d x + \int_{\frac{1}{2}}^{42} \frac{4}{x} - \frac{x}{8} d x

\int_{- 42}^{- \frac{1}{2}} \frac{4}{x} - \frac{x}{8} d x = 12 ln (2) - \frac{63}{32}

\int_{- \frac{1}{2}}^{0} \frac{x}{8} - 8 x d x = \frac{63}{32}

\int_{0}^{\frac{1}{2}} 8 x - \frac{x}{8} d x = \frac{63}{32}

\int_{\frac{1}{2}}^{42} \frac{4}{x} - \frac{x}{8} d x = 12 ln (2) - \frac{63}{32}

= 12 ln (2) - \frac{63}{32} + \frac{63}{32} + \frac{63}{32} + 12 ln (2) - \frac{63}{32}

Simplify

= 24 ln (2)

\frac{\partial}{\partial x} (\frac{3 x + t}{x - 5 t})

\int_{1}^{2} (8 x^{3} + 3 x^{2}) d x

\frac{d}{d x} (\frac{x ^{3}}{3} - x)

n = 2 \sum \infty - (\frac{2}{3})^{n}

x \to 0 + lim (\frac{ln ( 1 + 2 x )}{x})

What is area x= 4/y ,x= y/8 ,x=8y ?
The answer to area x= 4/y ,x= y/8 ,x=8y is 24ln(2)