What is area y= 1/x ,y=4x,y= 1/4 x ?

The answer to area y= 1/x ,y=4x,y= 1/4 x is 4ln(2)

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area y= 1/x ,y=4x,y= 1/4 x

area

y = \frac{1}{x}, y = 4 x, y = \frac{1}{4} x

4 ln (2)

Solution steps

To find the intersection points solve

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

\frac{1}{x} = 4 x : x = \frac{1}{2}, x = - \frac{1}{2}

\frac{1}{x} = \frac{1}{4} x : x = 2, x = - 2

4 x = \frac{1}{4} x : x = 0

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

= \int_{- 2}^{- \frac{1}{2}} \frac{1}{x} - \frac{1}{4} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{4} x - 4 x d x + \int_{0}^{\frac{1}{2}} 4 x - \frac{1}{4} x d x + \int_{\frac{1}{2}}^{2} \frac{1}{x} - \frac{1}{4} x d x

\int_{- 2}^{- \frac{1}{2}} \frac{1}{x} - \frac{1}{4} x d x = 2 ln (2) - \frac{15}{32}

\int_{- \frac{1}{2}}^{0} \frac{1}{4} x - 4 x d x = \frac{15}{32}

\int_{0}^{\frac{1}{2}} 4 x - \frac{1}{4} x d x = \frac{15}{32}

\int_{\frac{1}{2}}^{2} \frac{1}{x} - \frac{1}{4} x d x = 2 ln (2) - \frac{15}{32}

= 2 ln (2) - \frac{15}{32} + \frac{15}{32} + \frac{15}{32} + 2 ln (2) - \frac{15}{32}

Simplify

= 4 ln (2)

f (t) = t cos (t) - sin (t)

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x \to 0 + lim (x^{2 x})

What is area y= 1/x ,y=4x,y= 1/4 x ?
The answer to area y= 1/x ,y=4x,y= 1/4 x is 4ln(2)