What is area 2/x ,4x, 4/x ?

The answer to area 2/x ,4x, 4/x is 4ln(2)+diverges-2

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area 2/x ,4x, 4/x

area

\frac{2}{x}, 4 x, \frac{4}{x}

4 ln (2) + diverges - 2

Solution steps

To find the intersection points solve

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

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

\frac{2}{x} = \frac{4}{x} :

No Solution for

x \in R

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

= \int_{- 1}^{- \frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{2}}^{\frac{1}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{1} ∣ f_{3} (x) - f_{2} (x) ∣ d x

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

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

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

diverges

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

= - 1 + 2 ln (2) + diverges + 2 ln (2) - 1

Simplify

= 4 ln (2) + diverges - 2

5 x

x y^{^{'}} - 4 y = x^{6} e^{x}

x \to \infty lim (\frac{x ^{99}}{e ^{x}})

\int_{1}^{a} (\frac{1}{4 x}) d x

e^{s i n (x)} cos (x)

What is area 2/x ,4x, 4/x ?
The answer to area 2/x ,4x, 4/x is 4ln(2)+diverges-2