What is the integral of y^2sqrt(4-y) ?

The integral of y^2sqrt(4-y) is -2/7 (4-y)^{7/2}+16/5 (4-y)^{5/2}-32/3 (4-y)^{3/2}+C

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integral of y^2sqrt(4-y)

\int y^{2} 4 - y d y

- \frac{2}{7} (4 - y)^{\frac{7}{2}} + \frac{16}{5} (4 - y)^{\frac{5}{2}} - \frac{32}{3} (4 - y)^{\frac{3}{2}} + C

Solution steps

\int y^{2} 4 - y d y

Apply u-substitution

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

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

Expand

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

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

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

\int 16 u d u = \frac{32}{3} u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}})

Substitute back

u = 4 - y

= - (\frac{2}{7} (4 - y)^{\frac{7}{2}} - \frac{16}{5} (4 - y)^{\frac{5}{2}} + \frac{32}{3} (4 - y)^{\frac{3}{2}})

Simplify

= - \frac{2}{7} (4 - y)^{\frac{7}{2}} + \frac{16}{5} (4 - y)^{\frac{5}{2}} - \frac{32}{3} (4 - y)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{7} (4 - y)^{\frac{7}{2}} + \frac{16}{5} (4 - y)^{\frac{5}{2}} - \frac{32}{3} (4 - y)^{\frac{3}{2}} + C

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What is the integral of y^2sqrt(4-y) ?
The integral of y^2sqrt(4-y) is -2/7 (4-y)^{7/2}+16/5 (4-y)^{5/2}-32/3 (4-y)^{3/2}+C