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

The integral from 1 to t of xe^{-x} is -te^{-t}-e^{-t}+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 t of xe^{-x}

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

- t e^{- t} - e^{- t} + \frac{2}{e}

Solution steps

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

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{1}^{t}

= [- e^{- x} x - \int - e^{- x} d x]_{1}^{t}

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

= [- e^{- x} x - e^{- x}]_{1}^{t}

Compute the boundaries:

- t e^{- t} - e^{- t} + \frac{2}{e}

= - t e^{- t} - e^{- t} + \frac{2}{e}

\int_{0}^{1} x^{3} + x^{2} d x

\int_{1}^{x^{2}} e^{t} ln (t) d t

\int_{3}^{12} lo g_{10} (x - 2) d x

\int_{1}^{3} (3 x - 9) d x

\int_{0}^{π} cos^{4} (5 x) d x

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