What is area 1/(10x^2),x, x/(10) ?

The answer to area 1/(10x^2),x, x/(10) is (3\sqrt[3]{2}\sqrt[3]{5}(10-10^{2/3}))/(200)

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area 1/(10x^2),x, x/(10)

area

\frac{1}{10 x ^{2}}, x, \frac{x}{10}

\frac{3 3 2 3 5 ( 10 - 1 0 ^{\frac{2}{3}} )}{200}

Decimal

0.17316 \dots

Solution steps

To find the intersection points solve

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

\frac{1}{10 x ^{2}} = x : x = \frac{1}{3 10}

\frac{1}{10 x ^{2}} = \frac{x}{10} : x = 1

x = \frac{x}{10} : x = 0

= \int_{0}^{\frac{1}{3 10}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 10}}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 10}} x - \frac{x}{10} d x + \int_{\frac{1}{3 10}}^{1} \frac{1}{10 x ^{2}} - \frac{x}{10} d x

\int_{0}^{\frac{1}{3 10}} x - \frac{x}{10} d x = \frac{9}{20 \cdot 1 0 ^{\frac{2}{3}}}

\int_{\frac{1}{3 10}}^{1} \frac{1}{10 x ^{2}} - \frac{x}{10} d x = \frac{- 3 \cdot 1 0 ^{\frac{2}{3}} + 2 \cdot 1 0 ^{\frac{2}{3}} 3 10 + 1}{20 \cdot 1 0 ^{\frac{2}{3}}}

= \frac{9}{20 \cdot 1 0 ^{\frac{2}{3}}} + \frac{- 3 \cdot 1 0 ^{\frac{2}{3}} + 2 \cdot 1 0 ^{\frac{2}{3}} 3 10 + 1}{20 \cdot 1 0 ^{\frac{2}{3}}}

Simplify

= \frac{3 3 2 3 5 ( 10 - 1 0 ^{\frac{2}{3}} )}{200}

\frac{d y}{d x} = k y

\frac{\partial}{\partial x} (e^{x} \cdot cos (x))

9 (x^{6} - 2)^{5}

\frac{d}{d x} (\frac{x ^{2} - 1}{x + 1})

f (x) = 3 x^{2} + 6 x

What is area 1/(10x^2),x, x/(10) ?
The answer to area 1/(10x^2),x, x/(10) is (3\sqrt[3]{2}\sqrt[3]{5}(10-10^{2/3}))/(200)