What is the integral of y^2\sqrt[5]{2-4y^3} ?

The integral of y^2\sqrt[5]{2-4y^3} is -5/72 (2-4y^3)^{6/5}+C

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integral of y^2\sqrt[5]{2-4y^3}

- \frac{5}{72} (2 - 4 y^{3})^{\frac{6}{5}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= - \frac{1}{12} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 2 - 4 y^{3}

= - \frac{1}{12} \cdot \frac{5}{6} (2 - 4 y^{3})^{\frac{6}{5}}

Simplify

- \frac{1}{12} \cdot \frac{5}{6} (2 - 4 y^{3})^{\frac{6}{5}} : - \frac{5}{72} (2 - 4 y^{3})^{\frac{6}{5}}

= - \frac{5}{72} (2 - 4 y^{3})^{\frac{6}{5}}

Add a constant to the solution

= - \frac{5}{72} (2 - 4 y^{3})^{\frac{6}{5}} + C

x \to 0 lim (cos (\frac{4}{x}))

(- \frac{3}{4}, \frac{1}{4}), (\frac{1}{2}, - \frac{3}{2})

f (x) = x^{2} - 4, g (x) = 0, x = - 4, x = 3

\frac{\partial}{\partial x} (2 x - cos (y))

\frac{d}{d x} (\frac{sin ( x )}{1 - sin ( x )})

What is the integral of y^2\sqrt[5]{2-4y^3} ?
The integral of y^2\sqrt[5]{2-4y^3} is -5/72 (2-4y^3)^{6/5}+C