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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 Notation

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)

\int (\frac{1}{x} - \frac{5}{4 x ^{2} + 1}) d x

\int \frac{( 17 )}{( x - 1 ) ( x ^{2} + 16 )} d x

\int x (3 x^{2} - 2) d x

\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

\frac{\partial}{\partial x} (cos (2 x + y) - z)

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)