What is area (4/x),(16x),(1/4 x) ?

The answer to area (4/x),(16x),(1/4 x) is 24ln(2)

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area (4/x),(16x),(1/4 x)

area

(\frac{4}{x}), (16 x), (\frac{1}{4} x)

24 ln (2)

Decimal

16.63553 \dots

Solution steps

To find the intersection points solve

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

(\frac{4}{x}) = (16 x) : x = \frac{1}{2}, x = - \frac{1}{2}

(\frac{4}{x}) = (\frac{1}{4} x) : x = 4, x = - 4

(16 x) = (\frac{1}{4} 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{4}{x}) - (\frac{1}{4} x) d x + \int_{- \frac{1}{2}}^{0} (\frac{1}{4} x) - (16 x) d x + \int_{0}^{\frac{1}{2}} (16 x) - (\frac{1}{4} x) d x + \int_{\frac{1}{2}}^{4} (\frac{4}{x}) - (\frac{1}{4} x) d x

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

\int_{- \frac{1}{2}}^{0} (\frac{1}{4} x) - (16 x) d x = \frac{63}{32}

\int_{0}^{\frac{1}{2}} (16 x) - (\frac{1}{4} x) d x = \frac{63}{32}

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

What is area (4/x),(16x),(1/4 x) ?
The answer to area (4/x),(16x),(1/4 x) is 24ln(2)