What is area y= 3/x ,y=12x,y= 1/(12x) ?

The answer to area y= 3/x ,y=12x,y= 1/(12x) is -35/12+6ln(6)+diverges

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area y= 3/x ,y=12x,y= 1/(12x)

area

y = \frac{3}{x}, y = 12 x, y = \frac{1}{12 x}

- \frac{35}{12} + 6 ln (6) + diverges

Solution steps

To find the intersection points solve

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

\frac{3}{x} = 12 x : x = \frac{1}{2}, x = - \frac{1}{2}

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

No Solution for

x \in R

12 x = \frac{1}{12 x} : x = \frac{1}{12}, x = - \frac{1}{12}

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

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

\int_{- \frac{1}{2}}^{- \frac{1}{12}} \frac{3}{x} - 12 x d x = 3 ln (6) - \frac{35}{24}

\int_{- \frac{1}{12}}^{\frac{1}{12}} \frac{3}{x} - \frac{1}{12 x} d x =

diverges

\int_{\frac{1}{12}}^{\frac{1}{2}} 12 x - \frac{3}{x} d x = - \frac{35}{24} + 3 ln (6)

= 3 ln (6) - \frac{35}{24} + diverges - \frac{35}{24} + 3 ln (6)

Simplify

= - \frac{35}{12} + 6 ln (6) + diverges

\int \frac{2 x}{x ^{2} + 4 x + 12} d x

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

\frac{d}{d x} (- 9 cot (x) csc (x))

\frac{d}{d x} (ln (\frac{3 + e ^{x}}{3 - e ^{x}}))

\frac{d}{d x} (a x^{4} + b x^{3} + c x^{2})

What is area y= 3/x ,y=12x,y= 1/(12x) ?
The answer to area y= 3/x ,y=12x,y= 1/(12x) is -35/12+6ln(6)+diverges