What is the integral from 0 to t of 5xe^{-3x} ?

The integral from 0 to t of 5xe^{-3x} is 5/9 (-3te^{-3t}-e^{-3t}+1)

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integral from 0 to t of 5xe^{-3x}

\int_{0}^{t} 5 x e^{- 3 x} d x

\frac{5}{9} (- 3 t e^{- 3 t} - e^{- 3 t} + 1)

Solution steps

\int_{0}^{t} 5 x e^{- 3 x} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{t} e^{- 3 x} x d x

Apply u-substitution

: \int_{0}^{- 3 t} \frac{e ^{u} u}{9} d u

= 5 \cdot \int_{0}^{- 3 t} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 5 \cdot \frac{1}{9} \cdot \int_{0}^{- 3 t} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- 3 t}

= 5 \cdot \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{- 3 t}

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

= 5 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{- 3 t}

Simplify

5 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{- 3 t} : \frac{5}{9} [e^{u} u - e^{u}]_{0}^{- 3 t}

= \frac{5}{9} [e^{u} u - e^{u}]_{0}^{- 3 t}

Compute the boundaries:

- 3 t e^{- 3 t} - e^{- 3 t} + 1

= \frac{5}{9} (- 3 t e^{- 3 t} - e^{- 3 t} + 1)

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What is the integral from 0 to t of 5xe^{-3x} ?
The integral from 0 to t of 5xe^{-3x} is 5/9 (-3te^{-3t}-e^{-3t}+1)