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

The integral from 0 to 1 of (t+1)e^{-t} is -3/e+2

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

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

- \frac{3}{e} + 2

Decimal

0.89636 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Expand

- e^{- t} (t + 1) - e^{- t} : - e^{- t} t - 2 e^{- t}

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

Compute the boundaries:

- \frac{3}{e} + 2

= - \frac{3}{e} + 2

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What is the integral from 0 to 1 of (t+1)e^{-t} ?
The integral from 0 to 1 of (t+1)e^{-t} is -3/e+2