What is the integral from 0 to t of te^{-kt} ?

The integral from 0 to t of te^{-kt} is (-ke^{-kt}t-e^{-kt}+1)/(k^2)

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integral from 0 to t of te^{-kt}

\int_{0}^{t} t e^{- k t} d t

\frac{- k e ^{- k t} t - e ^{- k t} + 1}{k ^{2}}

Solution steps

\int_{0}^{t} t e^{- k t} d t

Apply u-substitution

: \int_{0}^{- k t} \frac{e ^{u} u}{k ^{2}} d u

= \int_{0}^{- k t} \frac{e ^{u} u}{k ^{2}} d u

Take the constant out:

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

= \frac{1}{k ^{2}} \cdot \int_{0}^{- k t} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- k t}

= \frac{1}{k ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- k t}

\int e^{u} d u = e^{u}

= \frac{1}{k ^{2}} [e^{u} u - e^{u}]_{0}^{- k t}

Simplify

\frac{1}{k ^{2}} [e^{u} u - e^{u}]_{0}^{- k t} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- k t}}{k ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- k t}}{k ^{2}}

Compute the boundaries:

- k e^{- k t} t - e^{- k t} + 1

= \frac{- k e ^{- k t} t - e ^{- k t} + 1}{k ^{2}}

\int_{0}^{x} \frac{x y}{1 + x ^{4}} d y

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\int_{1}^{3} (3 x - 2) d x

\int_{c o s (x)}^{π} (t + 1)^{2} d t

\int_{0}^{120} 3 - 3 e^{- 0.15 t} d t

What is the integral from 0 to t of te^{-kt} ?
The integral from 0 to t of te^{-kt} is (-ke^{-kt}t-e^{-kt}+1)/(k^2)