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

The answer to area y= 3/x ,y=9x,y= 1/9 x is 12ln(3)

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

area

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

12 ln (3)

Decimal

13.18334 \dots

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 : x = 33, x = - 33

9 x = \frac{1}{9} x : x = 0

= \int_{- 33}^{- \frac{1}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{3}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{1}{3}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3}}^{33} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 33}^{- \frac{1}{3}} \frac{3}{x} - \frac{1}{9} x d x + \int_{- \frac{1}{3}}^{0} \frac{1}{9} x - 9 x d x + \int_{0}^{\frac{1}{3}} 9 x - \frac{1}{9} x d x + \int_{\frac{1}{3}}^{33} \frac{3}{x} - \frac{1}{9} x d x

\int_{- 33}^{- \frac{1}{3}} \frac{3}{x} - \frac{1}{9} x d x = 6 ln (3) - \frac{40}{27}

\int_{- \frac{1}{3}}^{0} \frac{1}{9} x - 9 x d x = \frac{40}{27}

\int_{0}^{\frac{1}{3}} 9 x - \frac{1}{9} x d x = \frac{40}{27}

\int_{\frac{1}{3}}^{33} \frac{3}{x} - \frac{1}{9} x d x = 6 ln (3) - \frac{40}{27}

= 6 ln (3) - \frac{40}{27} + \frac{40}{27} + \frac{40}{27} + 6 ln (3) - \frac{40}{27}

Simplify

= 12 ln (3)

\int_{3}^{4} x x - 3 d x

f (x) = \frac{1}{x + 9}, at x = - 7

\frac{\partial}{\partial x} (3 y e^{4 x y} + 8 x)

\int \frac{3}{12 x - 1} d x

f (x) = \frac{x}{e ^{x}}

What is area y= 3/x ,y=9x,y= 1/9 x ?
The answer to area y= 3/x ,y=9x,y= 1/9 x is 12ln(3)