What is the integral from 0 to t of 20te^{-10t} ?

The integral from 0 to t of 20te^{-10t} is 1/5 (-10e^{-10t}t-e^{-10t}+1)

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integral from 0 to t of 20te^{-10t}

\int_{0}^{t} 20 t e^{- 10 t} d t

\frac{1}{5} (- 10 e^{- 10 t} t - e^{- 10 t} + 1)

Solution steps

\int_{0}^{t} 20 t e^{- 10 t} d t

Take the constant out:

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

= 20 \cdot \int_{0}^{t} t e^{- 10 t} d t

Apply u-substitution

: \int_{0}^{- 10 t} \frac{e ^{u} u}{100} d u

= 20 \cdot \int_{0}^{- 10 t} \frac{e ^{u} u}{100} d u

Take the constant out:

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

= 20 \cdot \frac{1}{100} \cdot \int_{0}^{- 10 t} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- 10 t}

= 20 \cdot \frac{1}{100} [e^{u} u - \int e^{u} d u]_{0}^{- 10 t}

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

= 20 \cdot \frac{1}{100} [e^{u} u - e^{u}]_{0}^{- 10 t}

Simplify

20 \cdot \frac{1}{100} [e^{u} u - e^{u}]_{0}^{- 10 t} : \frac{1}{5} [e^{u} u - e^{u}]_{0}^{- 10 t}

= \frac{1}{5} [e^{u} u - e^{u}]_{0}^{- 10 t}

Compute the boundaries:

- 10 e^{- 10 t} t - e^{- 10 t} + 1

= \frac{1}{5} (- 10 e^{- 10 t} t - e^{- 10 t} + 1)

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What is the integral from 0 to t of 20te^{-10t} ?
The integral from 0 to t of 20te^{-10t} is 1/5 (-10e^{-10t}t-e^{-10t}+1)