What is area y= 1/(12x^2),y=x,y= x/(18) ?

The answer to area y= 1/(12x^2),y=x,y= x/(18) is (3\sqrt[3]{12}-3^{2/3}\sqrt[3]{2})/(24)

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area y= 1/(12x^2),y=x,y= x/(18)

area

y = \frac{1}{12 x ^{2}}, y = x, y = \frac{x}{18}

\frac{3 3 12 - 3 ^{\frac{2}{3}} 3 2}{24}

Decimal

0.17698 \dots

Solution steps

To find the intersection points solve

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

\frac{1}{12 x ^{2}} = x : x = \frac{1}{3 12}

\frac{1}{12 x ^{2}} = \frac{x}{18} : x = 3 \frac{3}{2}

x = \frac{x}{18} : x = 0

= \int_{0}^{\frac{1}{3 12}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 12}}^{3 \frac{3}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 12}} x - \frac{x}{18} d x + \int_{\frac{1}{3 12}}^{3 \frac{3}{2}} \frac{1}{12 x ^{2}} - \frac{x}{18} d x

\int_{0}^{\frac{1}{3 12}} x - \frac{x}{18} d x = \frac{17}{36 \cdot 1 2 ^{\frac{2}{3}}}

\int_{\frac{1}{3 12}}^{3 \frac{3}{2}} \frac{1}{12 x ^{2}} - \frac{x}{18} d x = \frac{- 3 \frac{2}{3} + 3 12}{12} - \frac{6 3 6 - 2 ^{\frac{2}{3}}}{144 \cdot 3 ^{\frac{2}{3}}}

= \frac{17}{36 \cdot 1 2 ^{\frac{2}{3}}} + \frac{- 3 \frac{2}{3} + 3 12}{12} - \frac{6 3 6 - 2 ^{\frac{2}{3}}}{144 \cdot 3 ^{\frac{2}{3}}}

Simplify

= \frac{3 3 12 - 3 ^{\frac{2}{3}} 3 2}{24}

\int \frac{3 x ^{5} - 6 x ^{2} + 2 x - 5}{x ^{3}} d x

n = 0 \sum \infty e^{4 n}

\int_{2}^{- 1} 3^{x} d x

n = 2 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

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

What is area y= 1/(12x^2),y=x,y= x/(18) ?
The answer to area y= 1/(12x^2),y=x,y= x/(18) is (3\sqrt[3]{12}-3^{2/3}\sqrt[3]{2})/(24)