What is area 1/((16x^2)),x, x/6 ?

The answer to area 1/((16x^2)),x, x/6 is (6*3^{2/3}\sqrt[3]{6}-\sqrt[3]{2}*6^{2/3}-4*3^{2/3})/(96)

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area 1/((16x^2)),x, x/6

area

\frac{1}{( 16 x ^{2} )}, x, \frac{x}{6}

Decimal Notation

0.10622 \dots

Solution steps

To find the intersection points solve

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

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

Simplify

j = 0 \sum \infty \frac{7}{5 ^{j}}

\frac{d}{d x} (\frac{x}{cos ( x )})

\frac{\partial}{\partial x} (x e^{4 x + 2 y})

\int \frac{1}{6 + 2 x} d x

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

What is area 1/((16x^2)),x, x/6 ?
The answer to area 1/((16x^2)),x, x/6 is (6*3^{2/3}\sqrt[3]{6}-\sqrt[3]{2}*6^{2/3}-4*3^{2/3})/(96)