What is the integral from 0 to 5 of te^{-0.5t} ?

The integral from 0 to 5 of te^{-0.5t} is 2.85081…

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integral from 0 to 5 of te^{-0.5t}

\int_{0}^{5} t e^{- 0.5 t} d t

2.85081 \dots

Solution steps

\int_{0}^{5} t e^{- 0.5 t} d t

Apply u-substitution

: \int_{0}^{- 2.5} \frac{2}{0.5} e^{u} u d u

= \int_{0}^{- 2.5} \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_{- 2.5}^{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_{- 2.5}^{0} e^{u} u d u

Simplify

= - 4 \cdot \int_{- 2.5}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 2.5}^{0}

= - 4 [e^{u} u - \int e^{u} d u]_{- 2.5}^{0}

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

= - 4 [e^{u} u - e^{u}]_{- 2.5}^{0}

Compute the boundaries:

- 0.71270 \dots

= - 4 (- 0.71270 \dots)

Simplify

= 2.85081 \dots

\int_{2}^{5} (x^{2} - 4) d x

\int_{1}^{\infty} x^{- 5} d x

\int_{3}^{4} \frac{x}{( x ^{2} + 5 ) ^{3}} d x

\int_{1}^{2} (e^{t} + 4 t) d t

\int_{0}^{1} x^{15} + 1 5^{x} d x

What is the integral from 0 to 5 of te^{-0.5t} ?
The integral from 0 to 5 of te^{-0.5t} is 2.85081…