What is area y= 1/(18x^2),y=x,y= x/6 ?

The answer to area y= 1/(18x^2),y=x,y= x/6 is (2\sqrt[3]{18}-2^{2/3}\sqrt[3]{6})/(24)

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area y= 1/(18x^2),y=x,y= x/6

area

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{6}

Decimal Notation

0.09820 \dots

Solution steps

To find the intersection points solve

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

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

Simplify

\int \frac{1}{- x ^{2} - a + b} d x

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

\frac{1}{3 s ^{2}} - \frac{5}{2 s}

y = 3 x, y = x

\frac{d}{d x} (ln (x^{4} 8 - 6 x))

What is area y= 1/(18x^2),y=x,y= x/6 ?
The answer to area y= 1/(18x^2),y=x,y= x/6 is (2\sqrt[3]{18}-2^{2/3}\sqrt[3]{6})/(24)