What is the integral of y^3e^{-2y} ?

The integral of y^3e^{-2y} is 1/16 (-8e^{-2y}y^3-3(4e^{-2y}y^2-2(-2e^{-2y}y-e^{-2y})))+C

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integral of y^3e^{-2y}

\int y^{3} e^{- 2 y} d y

\frac{1}{16} (- 8 e^{- 2 y} y^{3} - 3 (4 e^{- 2 y} y^{2} - 2 (- 2 e^{- 2 y} y - e^{- 2 y}))) + C

Solution steps

\int y^{3} e^{- 2 y} d y

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{16} d u

= \int \frac{e ^{u} u ^{3}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= \frac{1}{16} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= \frac{1}{16} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 2 y

= \frac{1}{16} ((- 2 y)^{3} e^{- 2 y} - 3 ((- 2 y)^{2} e^{- 2 y} - 2 (e^{- 2 y} (- 2 y) - e^{- 2 y})))

Simplify

\frac{1}{16} ((- 2 y)^{3} e^{- 2 y} - 3 ((- 2 y)^{2} e^{- 2 y} - 2 (e^{- 2 y} (- 2 y) - e^{- 2 y}))) : \frac{1}{16} (- 8 e^{- 2 y} y^{3} - 3 (4 e^{- 2 y} y^{2} - 2 (- 2 e^{- 2 y} y - e^{- 2 y})))

= \frac{1}{16} (- 8 e^{- 2 y} y^{3} - 3 (4 e^{- 2 y} y^{2} - 2 (- 2 e^{- 2 y} y - e^{- 2 y})))

Add a constant to the solution

= \frac{1}{16} (- 8 e^{- 2 y} y^{3} - 3 (4 e^{- 2 y} y^{2} - 2 (- 2 e^{- 2 y} y - e^{- 2 y}))) + C

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What is the integral of y^3e^{-2y} ?
The integral of y^3e^{-2y} is 1/16 (-8e^{-2y}y^3-3(4e^{-2y}y^2-2(-2e^{-2y}y-e^{-2y})))+C