What is the integral from 0 to 5 of te^{-0.3t} ?

The integral from 0 to 5 of te^{-0.3t} is 4.91305…

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integral from 0 to 5 of te^{-0.3t}

\int_{0}^{5} t e^{- 0.3 t} d t

4.91305 \dots

Solution steps

\int_{0}^{5} t e^{- 0.3 t} d t

Apply u-substitution

: \int_{0}^{- 1.5} \frac{10 e ^{u} u}{0.9} d u

= \int_{0}^{- 1.5} \frac{10 e ^{u} u}{0.9} d u

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

= - \int_{- 1.5}^{0} \frac{10 e ^{u} u}{0.9} d u

Take the constant out:

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

= - \frac{10}{0.9} \cdot \int_{- 1.5}^{0} e^{u} u d u

Simplify

= - 11.11111 \dots \cdot \int_{- 1.5}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 1.5}^{0}

= - 11.11111 \dots [e^{u} u - \int e^{u} d u]_{- 1.5}^{0}

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

= - 11.11111 \dots [e^{u} u - e^{u}]_{- 1.5}^{0}

Compute the boundaries:

- 0.44217 \dots

= - 11.11111 \dots (- 0.44217 \dots)

Simplify

= 4.91305 \dots

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\int_{2}^{6} \frac{5}{x} - \frac{10}{x ^{2}} d x

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\int_{0}^{2} (4 - x) d x

What is the integral from 0 to 5 of te^{-0.3t} ?
The integral from 0 to 5 of te^{-0.3t} is 4.91305…