What is area 2/x ,8x, 1/(8x) ?

The answer to area 2/x ,8x, 1/(8x) is -15/8+8ln(2)+diverges

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area 2/x ,8x, 1/(8x)

area

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

- \frac{15}{8} + 8 ln (2) + diverges

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} :

No Solution for

x \in R

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

= \int_{- \frac{1}{2}}^{- \frac{1}{8}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{1}{8}}^{\frac{1}{8}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{8}}^{\frac{1}{2}} ∣ f_{2} (x) - f_{1} (x) ∣ d x

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

\int_{- \frac{1}{2}}^{- \frac{1}{8}} \frac{2}{x} - 8 x d x = 4 ln (2) - \frac{15}{16}

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

diverges

\int_{\frac{1}{8}}^{\frac{1}{2}} 8 x - \frac{2}{x} d x = - \frac{15}{16} + 4 ln (2)

= 4 ln (2) - \frac{15}{16} + diverges - \frac{15}{16} + 4 ln (2)

Simplify

= - \frac{15}{8} + 8 ln (2) + diverges

\int (3 x - 3) e^{3 x^{2} - 6 x} d x

y = 1 7^{x}

n = 1 \sum \infty \frac{( - 1 ) ^{3 n}}{n}

f (x) = \frac{6 x}{6 x ^{2} + 5}

\frac{\partial}{\partial x} (sin (π (6 x - 5 y)))

What is area 2/x ,8x, 1/(8x) ?
The answer to area 2/x ,8x, 1/(8x) is -15/8+8ln(2)+diverges