What is area x, 1/(18x^2), x/(16) ?

The answer to area x, 1/(18x^2), x/(16) is (\sqrt[3]{2}(4*3^{2/3}-6^{2/3}))/(48)

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area x, 1/(18x^2), x/(16)

area

x, \frac{1}{18 x ^{2}}, \frac{x}{16}

Decimal Notation

0.13172 \dots

Solution steps

To find the intersection points solve

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

x = \frac{x}{16} : x = 0

Simplify

\frac{d y}{d x} = 60 y x^{14}

\int \frac{40}{1 - cos ( 5 x )} d x

\int \frac{1}{7 x lo g _{7} ( x )} d x

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

e^{x} sin (10 x)

What is area x, 1/(18x^2), x/(16) ?
The answer to area x, 1/(18x^2), x/(16) is (\sqrt[3]{2}(4*3^{2/3}-6^{2/3}))/(48)