What is the integral from 0 to 4 of te^{-0.5t} ?

The integral from 0 to 4 of te^{-0.5t} is -4(-1+3/(e^2))

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integral from 0 to 4 of te^{-0.5t}

\int_{0}^{4} t e^{- 0.5 t} d t

- 4 (- 1 + \frac{3}{e ^{2}})

Decimal

2.37597 \dots

Solution steps

\int_{0}^{4} t e^{- 0.5 t} d t

Apply u-substitution

: \int_{0}^{- 2} \frac{2}{0.5} e^{u} u d u

= \int_{0}^{- 2} \frac{2}{0.5} e^{u} u d u

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

= - \int_{- 2}^{0} \frac{2}{0.5} e^{u} u d u

Take the constant out:

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

= - \frac{2}{0.5} \cdot \int_{- 2}^{0} e^{u} u d u

Simplify

= - 4 \cdot \int_{- 2}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 2}^{0}

= - 4 [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

= - 4 [e^{u} u - e^{u}]_{- 2}^{0}

Compute the boundaries:

- 1 + \frac{3}{e ^{2}}

= - 4 (- 1 + \frac{3}{e ^{2}})

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\int_{- 4}^{1} x d x

\int_{0}^{4} \frac{1}{x ^{2} + 2 x - 8} d x

What is the integral from 0 to 4 of te^{-0.5t} ?
The integral from 0 to 4 of te^{-0.5t} is -4(-1+3/(e^2))