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

The integral from 0 to infinity of te^{-3t} is 1/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{9}

Decimal Notation

0.11111 \dots

Solution steps

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

Compute the indefinite integral:

\int t e^{- 3 t} d t = \frac{1}{9} (- 3 e^{- 3 t} t - e^{- 3 t}) + C

Compute the boundaries:

\int_{0}^{\infty} t e^{- 3 t} d t = 0 - (- \frac{1}{9})

= 0 - (- \frac{1}{9})

Simplify

= \frac{1}{9}

x \to 6 lim (8 (x - 5) (x - 7))

\frac{\partial}{\partial x} (5 x y \cdot e^{- x^{2} y})

y = 4 x - x^{2}

\int \frac{8 x ^{3}}{3 y ^{2}}

\frac{\partial}{\partial x} (e^{y} + x)

What is the integral from 0 to infinity of te^{-3t} ?
The integral from 0 to infinity of te^{-3t} is 1/9