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

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

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

area

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

6 ln (12)

Decimal Notation

14.90943 \dots

Solution steps

To find the intersection points solve

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

\frac{3}{x} = \frac{1}{12} x : x = 6, x = - 6

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

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

= \int_{- 6}^{- \frac{1}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{1}{2}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{\frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 6}^{- \frac{1}{2}} \frac{3}{x} - \frac{1}{12} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{12} x - 12 x d x + \int_{0}^{\frac{1}{2}} \frac{1}{12} x - 12 x d x + \int_{\frac{1}{2}}^{6} \frac{3}{x} - \frac{1}{12} x d x

\int_{- 6}^{- \frac{1}{2}} \frac{3}{x} - \frac{1}{12} x d x = 3 ln (12) - \frac{143}{96}

\int_{- \frac{1}{2}}^{0} \frac{1}{12} x - 12 x d x = \frac{143}{96}

\int_{0}^{\frac{1}{2}} \frac{1}{12} x - 12 x d x = \frac{143}{96}

\int_{\frac{1}{2}}^{6} \frac{3}{x} - \frac{1}{12} x d x = 3 ln (12) - \frac{143}{96}

= 3 ln (12) - \frac{143}{96} + \frac{143}{96} + \frac{143}{96} + 3 ln (12) - \frac{143}{96}

Simplify

= 6 ln (12)

\frac{x - 4}{- x ^{2} + 4}

\frac{d}{d x} (tan (3 x - x^{2}))

\frac{d y}{d x} = \frac{1}{x sin ( y ) + 2 sin ( 2 y )}

\frac{\partial}{\partial x} (\frac{60}{1 + x ^{2} + y ^{2}})

\frac{d y}{d x} = 5 + y - 5 x + 2

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