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

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

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

area

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

6 ln (6)

Decimal Notation

10.75055 \dots

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}{3} x : x = 3, x = - 3

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

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

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

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

\int_{- \frac{1}{2}}^{0} \frac{1}{3} x - 12 x d x = \frac{35}{24}

\int_{0}^{\frac{1}{2}} 12 x - \frac{1}{3} x d x = \frac{35}{24}

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

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

Simplify

= 6 ln (6)

h \to 0 lim (\frac{( x - h ) - x}{h})

f (x) = sec (- 4 x^{2} - 3 x + 2)

\frac{d}{d x} (e^{- \frac{1}{x ^{7}}})

x \to 0 lim (tan (x^{2}) ln (x))

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