What is the integral from-infinity to 0 of xe^{-2x} ?

The integral from-infinity to 0 of xe^{-2x} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-infinity to 0 of xe^{-2x}

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

diverges

Solution steps

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

Compute the indefinite integral:

\int x e^{- 2 x} d x = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

Compute the boundaries:

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

diverges

= diverges

\frac{d}{d x} (\frac{2 e ^{x} - 4 x - 6}{5 sin ( x ) - 3})

\frac{\partial}{\partial x} (ln (x) tan (y))

\frac{\partial}{\partial x} (\frac{1}{x} E (x))

\int \frac{1}{( 4 + 9 x ^{2} ) \frac{5}{2}} d x

\int \frac{x}{3} d x

What is the integral from-infinity to 0 of xe^{-2x} ?
The integral from-infinity to 0 of xe^{-2x} is diverges