What is area y= 1/(4x^2),y=x,y= x/(16) ?

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

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area y= 1/(4x^2),y=x,y= x/(16)

area

y = \frac{1}{4 x ^{2}}, y = x, y = \frac{x}{16}

Solution steps

To find the intersection points solve

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

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

Simplify

\int (12 x + 14) (3 x^{2} + 7 x - 1)^{5} d x

n = 1 \sum \infty \frac{6}{n} - \frac{6}{n + 1}

4 y^{^{''}} - y = 0

y = \frac{1}{2} x^{2} + 9 x

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

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