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

The answer to area y= 9/x ,y=3x,y= 1/3 x is 18ln(3)

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

area

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

18 ln (3)

Decimal

19.77502 \dots

Solution steps

To find the intersection points solve

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

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

\frac{9}{x} = \frac{1}{3} x : x = 33, x = - 33

3 x = \frac{1}{3} x : x = 0

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

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

\int_{- 33}^{- 3} \frac{9}{x} - \frac{1}{3} x d x = 9 ln (3) - 4

\int_{- 3}^{0} \frac{1}{3} x - 3 x d x = 4

\int_{0}^{3} 3 x - \frac{1}{3} x d x = 4

\int_{3}^{33} \frac{9}{x} - \frac{1}{3} x d x = 9 ln (3) - 4

= 9 ln (3) - 4 + 4 + 4 + 9 ln (3) - 4

Simplify

= 18 ln (3)

\frac{d}{d x} (e^{- x} (cos (2 x) + sin (2 x)))

e^{- 2 t} cos (3 t) + 5 e^{- 2 t} sin (3 t)

x \to 2 - lim (\frac{x - 2}{x ^{2} - 4})

\frac{x ^{2} - 2}{x ^{2} + 2}

f (x) = 5 3 x

What is area y= 9/x ,y=3x,y= 1/3 x ?
The answer to area y= 9/x ,y=3x,y= 1/3 x is 18ln(3)