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

The answer to area 4/x ,16x, 1/16 x is 32ln(2)

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area 4/x ,16x, 1/16 x

area

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

32 ln (2)

Decimal Notation

22.18070 \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}{16} x : x = 8, x = - 8

16 x = \frac{1}{16} x : x = 0

= \int_{- 8}^{- \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}}^{8} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 8}^{- \frac{1}{2}} \frac{4}{x} - \frac{1}{16} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{16} x - 16 x d x + \int_{0}^{\frac{1}{2}} 16 x - \frac{1}{16} x d x + \int_{\frac{1}{2}}^{8} \frac{4}{x} - \frac{1}{16} x d x

\int_{- 8}^{- \frac{1}{2}} \frac{4}{x} - \frac{1}{16} x d x = 16 ln (2) - \frac{255}{128}

\int_{- \frac{1}{2}}^{0} \frac{1}{16} x - 16 x d x = \frac{255}{128}

\int_{0}^{\frac{1}{2}} 16 x - \frac{1}{16} x d x = \frac{255}{128}

\int_{\frac{1}{2}}^{8} \frac{4}{x} - \frac{1}{16} x d x = 16 ln (2) - \frac{255}{128}

= 16 ln (2) - \frac{255}{128} + \frac{255}{128} + \frac{255}{128} + 16 ln (2) - \frac{255}{128}

Simplify

= 32 ln (2)

\int x^{2} x + 4 d x

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\int \frac{x ^{2} + 1}{( x + 1 ) ^{2}} e^{x} d x

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