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

The answer to area y= 1/(16x^2),y=x,y= x/(18) is (\sqrt[3]{2}(18-18^{2/3}))/(96)

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

area

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

Decimal Notation

0.14609 \dots

Solution steps

To find the intersection points solve

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

x = \frac{x}{18} : x = 0

Simplify

f (x) = sin (cos (4 x))

\int 4 x cos (4 x) d x

\int (2^{x} - x^{2})^{2} d x

f (x) = 1.5 e^{0.76 x}

4 e^{(3 t)}

What is area y= 1/(16x^2),y=x,y= x/(18) ?
The answer to area y= 1/(16x^2),y=x,y= x/(18) is (\sqrt[3]{2}(18-18^{2/3}))/(96)