What is area y= 3/x ,y=9x,y= 1/(9x) ?

The answer to area y= 3/x ,y=9x,y= 1/(9x) is -26/9+9ln(3)+diverges

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area y= 3/x ,y=9x,y= 1/(9x)

area

y = \frac{3}{x}, y = 9 x, y = \frac{1}{9 x}

- \frac{26}{9} + 9 ln (3) + diverges

Solution steps

To find the intersection points solve

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

\frac{3}{x} = 9 x : x = \frac{1}{3}, x = - \frac{1}{3}

\frac{3}{x} = \frac{1}{9 x} :

No Solution for

x \in R

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

= \int_{- \frac{1}{3}}^{- \frac{1}{9}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{1}{9}}^{\frac{1}{9}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{9}}^{\frac{1}{3}} ∣ f_{2} (x) - f_{1} (x) ∣ d x

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

\int_{- \frac{1}{3}}^{- \frac{1}{9}} \frac{3}{x} - 9 x d x = \frac{9}{2} ln (3) - \frac{13}{9}

\int_{- \frac{1}{9}}^{\frac{1}{9}} \frac{3}{x} - \frac{1}{9 x} d x =

diverges

\int_{\frac{1}{9}}^{\frac{1}{3}} 9 x - \frac{3}{x} d x = - \frac{13}{9} + \frac{9}{2} ln (3)

= \frac{9}{2} ln (3) - \frac{13}{9} + diverges - \frac{13}{9} + \frac{9}{2} ln (3)

Simplify

= - \frac{26}{9} + 9 ln (3) + diverges

4 sin (3 t)

\frac{\partial}{\partial x} (sin (x^{2} + 2 y))

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) d x

\int \frac{22}{22 + e ^{x}} d x

f (x) = 9 - x^{2}, (3, 0)

What is area y= 3/x ,y=9x,y= 1/(9x) ?
The answer to area y= 3/x ,y=9x,y= 1/(9x) is -26/9+9ln(3)+diverges