What is the integral from 0 to 5 of 0.05te^{-0.35t} ?

The integral from 0 to 5 of 0.05te^{-0.35t} is 0.21311…

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integral from 0 to 5 of 0.05te^{-0.35t}

\int_{0}^{5} 0.05 t e^{- 0.35 t} d t

0.21311 \dots

Solution steps

\int_{0}^{5} 0.05 t e^{- 0.35 t} d t

Take the constant out:

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

= 0.05 \cdot \int_{0}^{5} t e^{- 0.35 t} d t

Apply u-substitution

: \int_{0}^{- 1.75} \frac{20 e ^{u} u}{2.45} d u

= 0.05 \cdot \int_{0}^{- 1.75} \frac{20 e ^{u} u}{2.45} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 0.05 (- \int_{- 1.75}^{0} \frac{20 e ^{u} u}{2.45} d u)

Take the constant out:

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

= 0.05 (- \frac{20}{2.45} \cdot \int_{- 1.75}^{0} e^{u} u d u)

Simplify

= 0.05 (- 8.16326 \dots \cdot \int_{- 1.75}^{0} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 1.75}^{0}

= 0.05 (- 8.16326 \dots [e^{u} u - \int e^{u} d u]_{- 1.75}^{0})

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

= 0.05 (- 8.16326 \dots [e^{u} u - e^{u}]_{- 1.75}^{0})

Simplify

= - 0.40816 \dots [e^{u} u - e^{u}]_{- 1.75}^{0}

Compute the boundaries:

- 0.52212 \dots

= - 0.40816 \dots (- 0.52212 \dots)

Simplify

= 0.21311 \dots

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What is the integral from 0 to 5 of 0.05te^{-0.35t} ?
The integral from 0 to 5 of 0.05te^{-0.35t} is 0.21311…