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

The answer to area y= 1/(8x^2),y=x,y= x/8 is 3/16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y= 1/(8x^2),y=x,y= x/8

area

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

\frac{3}{16}

Decimal Notation

0.1875

Solution steps

To find the intersection points solve

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

\frac{1}{8 x ^{2}} = x : x = \frac{1}{2}

\frac{1}{8 x ^{2}} = \frac{x}{8} : x = 1

x = \frac{x}{8} : x = 0

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

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

\int_{0}^{\frac{1}{2}} x - \frac{x}{8} d x = \frac{7}{64}

\int_{\frac{1}{2}}^{1} \frac{1}{8 x ^{2}} - \frac{x}{8} d x = \frac{5}{64}

= \frac{7}{64} + \frac{5}{64}

Simplify

= \frac{3}{16}

\frac{d}{d x} (x sec (y))

\frac{1}{x - 4}

\int_{0}^{4} (x - (x - 2)) d x

y = \frac{- 10}{x ^{4}}

\frac{d}{d t} (2 t)

What is area y= 1/(8x^2),y=x,y= x/8 ?
The answer to area y= 1/(8x^2),y=x,y= x/8 is 3/16