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Popular Calculus >

area 1/8 x,8x, 2/x

Solution

area

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

Solution

12 ln (2)

+1

Decimal

8.31776 \dots

Solution steps

To find the intersection points solve

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

\frac{1}{8} x = 8 x : x = 0

\frac{1}{8} x = \frac{2}{x} : x = 4, x = - 4

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

= \int_{- 4}^{- \frac{1}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{\frac{1}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{1}{2}}^{4} ∣ f_{1} (x) - f_{3} (x) ∣ d x

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

\int_{- 4}^{- \frac{1}{2}} \frac{1}{8} x - \frac{2}{x} d x = - \frac{63}{64} + 6 ln (2)

\int_{- \frac{1}{2}}^{0} 8 x - \frac{1}{8} x d x = \frac{63}{64}

\int_{0}^{\frac{1}{2}} \frac{1}{8} x - 8 x d x = \frac{63}{64}

\int_{\frac{1}{2}}^{4} \frac{1}{8} x - \frac{2}{x} d x = - \frac{63}{64} + 6 ln (2)

= - \frac{63}{64} + 6 ln (2) + \frac{63}{64} + \frac{63}{64} - \frac{63}{64} + 6 ln (2)

Simplify

= 12 ln (2)

Graph

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Frequently Asked Questions (FAQ)

What is area 1/8 x,8x, 2/x ?
The answer to area 1/8 x,8x, 2/x is 12ln(2)