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

The answer to area y= 1/(x^2),y=x,y= 1/8 x is 3/4

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

area

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

\frac{3}{4}

Decimal

0.75

Solution steps

To find the intersection points solve

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

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

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

x = \frac{1}{8} x : x = 0

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

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

\int_{0}^{1} x - \frac{1}{8} x d x = \frac{7}{16}

\int_{1}^{2} \frac{1}{x ^{2}} - \frac{1}{8} x d x = \frac{5}{16}

= \frac{7}{16} + \frac{5}{16}

Simplify

= \frac{3}{4}

\frac{d}{d x} (50 e^{- 0.5 x} - (\frac{5}{x}))

\frac{\partial}{\partial x} (x + e^{4 y})

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

- (x)^{2} 1

\frac{d}{d t} (- e^{3 t})

What is area y= 1/(x^2),y=x,y= 1/8 x ?
The answer to area y= 1/(x^2),y=x,y= 1/8 x is 3/4