What is area y= 1/(8x^2),y=x,y= x/6 ?

The answer to area y= 1/(8x^2),y=x,y= x/6 is 5/48+(-2\sqrt[3]{36}+13-\sqrt[3]{4}*3^{2/3})/(48)

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area y= 1/(8x^2),y=x,y= x/6

area

y = \frac{1}{8 x ^{2}}, y = x, y = \frac{x}{6}

Decimal Notation

0.16862 \dots

Solution steps

To find the intersection points solve

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

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

x = \frac{x}{6} : x = 0

\int_{0}^{\frac{1}{2}} x - \frac{x}{6} d x = \frac{5}{48}

Simplify

\frac{\partial}{\partial z} (2 x y z^{- 1})

x^{2} + 1, 2

\int \frac{3 x ^{2}}{1 - x ^{3}} d x

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

y^{2} d x + (2 x y + 1) d y = 0, y (9) = 90

What is area y= 1/(8x^2),y=x,y= x/6 ?
The answer to area y= 1/(8x^2),y=x,y= x/6 is 5/48+(-2\sqrt[3]{36}+13-\sqrt[3]{4}*3^{2/3})/(48)