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

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

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

area

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

To find the intersection points solve

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

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

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

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

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

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

\int_{- 2}^{- 1} \frac{1}{x} - \frac{1}{4} x d x = ln (2) - \frac{3}{8}

\int_{- 1}^{0} \frac{1}{4} x - x d x = \frac{3}{8}

\int_{0}^{1} x - \frac{1}{4} x d x = \frac{3}{8}

\int_{1}^{2} \frac{1}{x} - \frac{1}{4} x d x = ln (2) - \frac{3}{8}

= ln (2) - \frac{3}{8} + \frac{3}{8} + \frac{3}{8} + ln (2) - \frac{3}{8}

Simplify

= 2 ln (2)

\int_{1}^{6} \frac{x ^{2} + 5}{7 x - x ^{2}} d x

\frac{cos ( 7 t )}{e ^{5 t}}

x \frac{d y}{d x} + y = y^{2} x^{2} ln (x)

\int \frac{y}{x ^{2} + y ^{2}} d x

\frac{2 s + 12}{( s ^{2} + 2 s + 5 )}

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