What is the integral of t^2e^{-20t} ?

The integral of t^2e^{-20t} is -1/4000 e^{-20t}(200t^2+20t+1)+C

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integral of t^2e^{-20t}

\int t^{2} e^{- 20 t} d t

- \frac{1}{4000} e^{- 20 t} (200 t^{2} + 20 t + 1) + C

Solution steps

\int t^{2} e^{- 20 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{8000} d u

= \int - \frac{e ^{u} u ^{2}}{8000} d u

Take the constant out:

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

= - \frac{1}{8000} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{8000} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{8000} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 20 t

= - \frac{1}{8000} ((- 20 t)^{2} e^{- 20 t} - 2 (e^{- 20 t} (- 20 t) - e^{- 20 t}))

Simplify

- \frac{1}{8000} ((- 20 t)^{2} e^{- 20 t} - 2 (e^{- 20 t} (- 20 t) - e^{- 20 t})) : - \frac{1}{4000} e^{- 20 t} (200 t^{2} + 20 t + 1)

= - \frac{1}{4000} e^{- 20 t} (200 t^{2} + 20 t + 1)

Add a constant to the solution

= - \frac{1}{4000} e^{- 20 t} (200 t^{2} + 20 t + 1) + C

\int_{t}^{1} ln (x) d x

\frac{1}{s ^{2} + 2}

\int \frac{x ^{2}}{x ^{2} - 2 x + 1} d x

\frac{1}{8 x ^{2}}

n = 1 \sum \infty 3^{n}

What is the integral of t^2e^{-20t} ?
The integral of t^2e^{-20t} is -1/4000 e^{-20t}(200t^2+20t+1)+C