What is area y= 4/x ,y=8x,y= 1/8 x ?

The answer to area y= 4/x ,y=8x,y= 1/8 x is 24ln(2)

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area y= 4/x ,y=8x,y= 1/8 x

area

y = \frac{4}{x}, y = 8 x, y = \frac{1}{8} x

24 ln (2)

Decimal Notation

16.63553 \dots

Solution steps

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{1}{8} x : x = 42, x = - 42

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

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

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

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

\int_{\frac{1}{2}}^{42} \frac{4}{x} - \frac{1}{8} x 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)

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What is area y= 4/x ,y=8x,y= 1/8 x ?
The answer to area y= 4/x ,y=8x,y= 1/8 x is 24ln(2)