What is the integral from 0 to 3 of te^{-0.5t} ?

The integral from 0 to 3 of te^{-0.5t} is 1.76869…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of te^{-0.5t}

\int_{0}^{3} t e^{- 0.5 t} d t

1.76869 \dots

Solution steps

\int_{0}^{3} t e^{- 0.5 t} d t

Apply u-substitution

: \int_{0}^{- 1.5} \frac{2}{0.5} e^{u} u d u

= \int_{0}^{- 1.5} \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_{- 1.5}^{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_{- 1.5}^{0} e^{u} u d u

Simplify

= - 4 \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}

= - 4 [e^{u} u - \int e^{u} d u]_{- 1.5}^{0}

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

= - 4 [e^{u} u - e^{u}]_{- 1.5}^{0}

Compute the boundaries:

- 0.44217 \dots

= - 4 (- 0.44217 \dots)

Simplify

= 1.76869 \dots

\int_{0}^{e^{x}} cos (t) d t

\int_{0}^{9} 1000 x + 250 d x

\int_{0}^{25} 3 x d x

\int_{9}^{16} (x - 4) d x

\int_{0}^{1} e^{3 x - 2} d x

What is the integral from 0 to 3 of te^{-0.5t} ?
The integral from 0 to 3 of te^{-0.5t} is 1.76869…