What is the integral from 1 to infinity of-xe^{-x} ?

The integral from 1 to infinity of-xe^{-x} is -2/e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of-xe^{-x}

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

- \frac{2}{e}

Decimal Notation

- 0.73575 \dots

Solution steps

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

Compute the indefinite integral:

\int - x e^{- x} d x = e^{- x} x + e^{- x} + C

Compute the boundaries:

\int_{1}^{\infty} - x e^{- x} d x = 0 - \frac{2}{e}

= 0 - \frac{2}{e}

Simplify

= - \frac{2}{e}

\int_{0}^{1} 12 x (x - x^{2}) d x

\int_{2}^{4} \frac{x - 2}{x} d x

\int_{- 1}^{0} (2 t + 2)^{2} d t

\int_{1}^{\infty} x e^{- a x} d x

\int_{0}^{\infty} x + 3 x^{2} d x

What is the integral from 1 to infinity of-xe^{-x} ?
The integral from 1 to infinity of-xe^{-x} is -2/e