What is area y= 2/x ,y=8x,y= 1/8 x ?

The answer to area y= 2/x ,y=8x,y= 1/8 x is 12ln(2)

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area y= 2/x ,y=8x,y= 1/8 x

area

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8} x

12 ln (2)

Decimal Notation

8.31776 \dots

Solution steps

To find the intersection points solve

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

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

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

8 x = \frac{1}{8} x : x = 0

= \int_{- 4}^{- \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}}^{4} ∣ f_{1} (x) - f_{3} (x) ∣ d x

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

\int_{- 4}^{- \frac{1}{2}} \frac{2}{x} - \frac{1}{8} x d x = 6 ln (2) - \frac{63}{64}

\int_{- \frac{1}{2}}^{0} \frac{1}{8} x - 8 x d x = \frac{63}{64}

\int_{0}^{\frac{1}{2}} 8 x - \frac{1}{8} x d x = \frac{63}{64}

\int_{\frac{1}{2}}^{4} \frac{2}{x} - \frac{1}{8} x d x = 6 ln (2) - \frac{63}{64}

= 6 ln (2) - \frac{63}{64} + \frac{63}{64} + \frac{63}{64} + 6 ln (2) - \frac{63}{64}

Simplify

= 12 ln (2)

\int_{2}^{\infty} \frac{1}{ln ( x )} d x

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\int_{1}^{2} cos (\frac{π x}{3}) d x

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What is area y= 2/x ,y=8x,y= 1/8 x ?
The answer to area y= 2/x ,y=8x,y= 1/8 x is 12ln(2)