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Popular >

area 9/x ,2x, 1/2 x

Solution

area

\frac{9}{x}, 2 x, \frac{1}{2} x

Solution

18 ln (2)

+1

Decimal Notation

12.47664 \dots

Solution steps

To find the intersection points solve

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

\frac{9}{x} = 2 x : x = \frac{3}{2}, x = - \frac{3}{2}

\frac{9}{x} = \frac{1}{2} x : x = 32, x = - 32

2 x = \frac{1}{2} x : x = 0

= \int_{- 32}^{- \frac{3}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{3}{2}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{3}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{3}{2}}^{32} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 32}^{- \frac{3}{2}} \frac{9}{x} - \frac{1}{2} x d x + \int_{- \frac{3}{2}}^{0} \frac{1}{2} x - 2 x d x + \int_{0}^{\frac{3}{2}} 2 x - \frac{1}{2} x d x + \int_{\frac{3}{2}}^{32} \frac{9}{x} - \frac{1}{2} x d x

\int_{- 32}^{- \frac{3}{2}} \frac{9}{x} - \frac{1}{2} x d x = 9 ln (2) - \frac{27}{8}

\int_{- \frac{3}{2}}^{0} \frac{1}{2} x - 2 x d x = \frac{27}{8}

\int_{0}^{\frac{3}{2}} 2 x - \frac{1}{2} x d x = \frac{27}{8}

\int_{\frac{3}{2}}^{32} \frac{9}{x} - \frac{1}{2} x d x = 9 ln (2) - \frac{27}{8}

= 9 ln (2) - \frac{27}{8} + \frac{27}{8} + \frac{27}{8} + 9 ln (2) - \frac{27}{8}

Simplify

= 18 ln (2)

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Frequently Asked Questions (FAQ)

What is area 9/x ,2x, 1/2 x ?
The answer to area 9/x ,2x, 1/2 x is 18ln(2)