What is the integral of y^2(4-y)^{2/3} ?

The integral of y^2(4-y)^{2/3} is -3/11 (4-y)^{11/3}+3(4-y)^{8/3}-48/5 (4-y)^{5/3}+C

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integral of y^2(4-y)^{2/3}

\int y^{2} (4 - y)^{\frac{2}{3}} d y

- \frac{3}{11} (4 - y)^{\frac{11}{3}} + 3 (4 - y)^{\frac{8}{3}} - \frac{48}{5} (4 - y)^{\frac{5}{3}} + C

Solution steps

\int y^{2} (4 - y)^{\frac{2}{3}} d y

Apply u-substitution

: \int - (- u + 4)^{2} u^{\frac{2}{3}} d u

= \int - (- u + 4)^{2} u^{\frac{2}{3}} d u

Take the constant out:

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

= - \int (- u + 4)^{2} u^{\frac{2}{3}} d u

Expand

(- u + 4)^{2} u^{\frac{2}{3}} : u^{\frac{8}{3}} - 8 u^{\frac{5}{3}} + 16 u^{\frac{2}{3}}

= - \int u^{\frac{8}{3}} - 8 u^{\frac{5}{3}} + 16 u^{\frac{2}{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u^{\frac{8}{3}} d u - \int 8 u^{\frac{5}{3}} d u + \int 16 u^{\frac{2}{3}} d u)

\int u^{\frac{8}{3}} d u = \frac{3}{11} u^{\frac{11}{3}}

\int 8 u^{\frac{5}{3}} d u = 3 u^{\frac{8}{3}}

\int 16 u^{\frac{2}{3}} d u = \frac{48}{5} u^{\frac{5}{3}}

= - (\frac{3}{11} u^{\frac{11}{3}} - 3 u^{\frac{8}{3}} + \frac{48}{5} u^{\frac{5}{3}})

Substitute back

u = 4 - y

= - (\frac{3}{11} (4 - y)^{\frac{11}{3}} - 3 (4 - y)^{\frac{8}{3}} + \frac{48}{5} (4 - y)^{\frac{5}{3}})

Simplify

= - \frac{3}{11} (4 - y)^{\frac{11}{3}} + 3 (4 - y)^{\frac{8}{3}} - \frac{48}{5} (4 - y)^{\frac{5}{3}}

Add a constant to the solution

= - \frac{3}{11} (4 - y)^{\frac{11}{3}} + 3 (4 - y)^{\frac{8}{3}} - \frac{48}{5} (4 - y)^{\frac{5}{3}} + C

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What is the integral of y^2(4-y)^{2/3} ?
The integral of y^2(4-y)^{2/3} is -3/11 (4-y)^{11/3}+3(4-y)^{8/3}-48/5 (4-y)^{5/3}+C