What is area y= 3/x ,y=7x,y= 1/7 x ?

The answer to area y= 3/x ,y=7x,y= 1/7 x is 6ln(7)

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area y= 3/x ,y=7x,y= 1/7 x

area

y = \frac{3}{x}, y = 7 x, y = \frac{1}{7} x

6 ln (7)

Decimal Notation

11.67546 \dots

Solution steps

To find the intersection points solve

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

\frac{3}{x} = 7 x : x = \frac{3}{7}, x = - \frac{3}{7}

\frac{3}{x} = \frac{1}{7} x : x = 21, x = - 21

7 x = \frac{1}{7} x : x = 0

= \int_{- 21}^{- \frac{3}{7}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{3}{7}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{3}{7}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{3}{7}}^{21} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 21}^{- \frac{3}{7}} \frac{3}{x} - \frac{1}{7} x d x + \int_{- \frac{3}{7}}^{0} \frac{1}{7} x - 7 x d x + \int_{0}^{\frac{3}{7}} 7 x - \frac{1}{7} x d x + \int_{\frac{3}{7}}^{21} \frac{3}{x} - \frac{1}{7} x d x

\int_{- 21}^{- \frac{3}{7}} \frac{3}{x} - \frac{1}{7} x d x = 3 ln (7) - \frac{72}{49}

\int_{- \frac{3}{7}}^{0} \frac{1}{7} x - 7 x d x = \frac{72}{49}

\int_{0}^{\frac{3}{7}} 7 x - \frac{1}{7} x d x = \frac{72}{49}

\int_{\frac{3}{7}}^{21} \frac{3}{x} - \frac{1}{7} x d x = 3 ln (7) - \frac{72}{49}

= 3 ln (7) - \frac{72}{49} + \frac{72}{49} + \frac{72}{49} + 3 ln (7) - \frac{72}{49}

Simplify

= 6 ln (7)

f (x) = \frac{( x ^{2} - 6 x )}{x + 1}

\frac{d}{d x} (e^{\frac{x}{a}})

\int x e^{- \frac{1}{2} x} d x

x \to 2 lim (1 0^{x^{2} - 4 x})

\int e x^{e + 1} d x

What is area y= 3/x ,y=7x,y= 1/7 x ?
The answer to area y= 3/x ,y=7x,y= 1/7 x is 6ln(7)