What is the integral from 0 to 6 of te^{-t} ?

The integral from 0 to 6 of te^{-t} is -7/(e^6)+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 6 of te^{-t}

\int_{0}^{6} t e^{- t} d t

- \frac{7}{e ^{6}} + 1

Solution steps

\int_{0}^{6} t e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t - \int - e^{- t} d t]_{0}^{6}

= [- e^{- t} t - \int - e^{- t} d t]_{0}^{6}

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

= [- e^{- t} t - e^{- t}]_{0}^{6}

Compute the boundaries:

- \frac{7}{e ^{6}} + 1

= - \frac{7}{e ^{6}} + 1

\frac{\partial}{\partial x} (w e^{x^{2} + y^{2}} ln (z))

\int \frac{y}{25 - 36 y ^{4}} d y

\int \frac{1}{x ^{2} ( 1 - x ^{2} )} d x

\frac{d}{d x} (\frac{( x + 1 )}{3})

\frac{d}{d x} (\frac{x ^{2}}{sin ( a ) x})

What is the integral from 0 to 6 of te^{-t} ?
The integral from 0 to 6 of te^{-t} is -7/(e^6)+1