What is area y^2=x,\sqrt[6]{y}=x ?

The answer to area y^2=x,\sqrt[6]{y}=x is 11/21

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area y^2=x,\sqrt[6]{y}=x

area

\frac{11}{21}

Solution steps

Isolate

y

for

y^{2} = x : y = x, y = - x

Isolate

y

for

f_{1} (x) = x

f_{2} (x) = - x

To find the intersection points solve

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

x = - x : x = 0

x = x^{6} : x = 0, x = 1

- x = x^{6} : x = 0

= \int_{0}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{1} x - x^{6} d x

\int_{0}^{1} x - x^{6} d x = \frac{11}{21}

= \frac{11}{21}

f (x) = x + 1, (8, 3)

\int \frac{4 x ^{2} - 64}{x} d x

y = ∣ x - 3 ∣, y = \frac{x}{2}, x = 0

\frac{d}{d x} (ln (\frac{e ^{7 x}}{1 + e ^{7 x}}))

3 tan^{2} (2 x)

What is area y^2=x,\sqrt[6]{y}=x ?
The answer to area y^2=x,\sqrt[6]{y}=x is 11/21