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

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

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area y= 4/x ,y=16x,y= 1/16 x

area

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

32 ln (2)

Decimal

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)

y = 3 x ln (x)

\frac{\partial}{\partial x} (\frac{2 x - 2 y}{2 x + 2 y})

f (x) = \frac{9}{x}, at x = 2

\int_{1}^{5} x^{2} d x

\frac{d ^{2} y}{d x ^{2}} + y = 1 + x

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