What is area y= 1/(20x^2),x=y,y= x/(16) ?

The answer to area y= 1/(20x^2),x=y,y= x/(16) is (3(2\sqrt[3]{5}*2^{2/3}-\sqrt[3]{10}))/(80)

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area y= 1/(20x^2),x=y,y= x/(16)

area

y = \frac{1}{20 x ^{2}}, x = y, y = \frac{x}{16}

Decimal Notation

0.12279 \dots

Solution steps

Isolate

y

for

x = y : y = x

f_{1} (x) = \frac{1}{20 x ^{2}}

f_{2} (x) = x

To find the intersection points solve

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

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

Simplify

\int x \cdot y

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

\frac{d}{d x} ((3 x^{4} - 18 x)^{5} \cdot cos (5 x^{2}))

\int (2 x^{2} + 3)^{2} d x

(y x - y) d y - (y + 1) d x = 0

What is area y= 1/(20x^2),x=y,y= x/(16) ?
The answer to area y= 1/(20x^2),x=y,y= x/(16) is (3(2\sqrt[3]{5}*2^{2/3}-\sqrt[3]{10}))/(80)