What is the integral from 0 to 1 of (2t+5)(e^{-t}) ?

The integral from 0 to 1 of (2t+5)(e^{-t}) is -9/e+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of (2t+5)(e^{-t})

\int_{0}^{1} (2 t + 5) (e^{- t}) d t

- \frac{9}{e} + 7

Decimal

3.68908 \dots

Solution steps

\int_{0}^{1} (2 t + 5) e^{- t} d t

Apply Integration By Parts

= [- e^{- t} (2 t + 5) - \int - 2 e^{- t} d t]_{0}^{1}

\int - 2 e^{- t} d t = 2 e^{- t}

= [- e^{- t} (2 t + 5) - 2 e^{- t}]_{0}^{1}

Expand

- e^{- t} (2 t + 5) - 2 e^{- t} : - 2 e^{- t} t - 7 e^{- t}

= [- 2 e^{- t} t - 7 e^{- t}]_{0}^{1}

Compute the boundaries:

- \frac{9}{e} + 7

= - \frac{9}{e} + 7

What is the integral from 0 to 1 of (2t+5)(e^{-t}) ?
The integral from 0 to 1 of (2t+5)(e^{-t}) is -9/e+7