What is area y= 2/x ,y=8x,y= 1/2 x ?

The answer to area y= 2/x ,y=8x,y= 1/2 x is 8ln(2)

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area y= 2/x ,y=8x,y= 1/2 x

area

y = \frac{2}{x}, y = 8 x, y = \frac{1}{2} x

8 ln (2)

Decimal

5.54517 \dots

Solution steps

To find the intersection points solve

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

\frac{2}{x} = 8 x : x = \frac{1}{2}, x = - \frac{1}{2}

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

8 x = \frac{1}{2} 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{2}{x} - \frac{1}{2} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{2} x - 8 x d x + \int_{0}^{\frac{1}{2}} 8 x - \frac{1}{2} x d x + \int_{\frac{1}{2}}^{2} \frac{2}{x} - \frac{1}{2} x d x

\int_{- 2}^{- \frac{1}{2}} \frac{2}{x} - \frac{1}{2} x d x = 4 ln (2) - \frac{15}{16}

\int_{- \frac{1}{2}}^{0} \frac{1}{2} x - 8 x d x = \frac{15}{16}

\int_{0}^{\frac{1}{2}} 8 x - \frac{1}{2} x d x = \frac{15}{16}

\int_{\frac{1}{2}}^{2} \frac{2}{x} - \frac{1}{2} x d x = 4 ln (2) - \frac{15}{16}

= 4 ln (2) - \frac{15}{16} + \frac{15}{16} + \frac{15}{16} + 4 ln (2) - \frac{15}{16}

Simplify

= 8 ln (2)

\frac{d}{d x} (\frac{3}{4 ( 2 x - 5 ) ^{8}})

x \to - \infty lim (\frac{x ^{2} + 8 x}{8 x + 1})

\int ln (x) e^{- x} d x

\int \frac{4 x ^{2} + 3}{( x - 5 ) ^{2}} d x

\frac{d}{d x} (\frac{( x + 5 ) ( x ^{2} - 5 x + 25 )}{x ^{3}})

What is area y= 2/x ,y=8x,y= 1/2 x ?
The answer to area y= 2/x ,y=8x,y= 1/2 x is 8ln(2)