What is the integral of (t^3)(e^{-2t}) ?

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

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integral of (t^3)(e^{-2t})

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

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

Solution steps

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

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 t

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = x^{2} - y - 2

2 x, \frac{2}{x}, x = 3

\frac{\partial}{\partial y} (4 cos (x^{2} + y^{2}))

x \to 0 lim (\frac{ln ( 1 + 5 x )}{x})

y^{^{'}} - \frac{y}{x} = 0

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