What is the integral from 0 to 1 of te^{-t} ?

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

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integral from 0 to 1 of te^{-t}

\int_{0}^{1} t e^{- t} d t

- \frac{2}{e} + 1

Decimal

0.26424 \dots

Solution steps

\int_{0}^{1} t e^{- t} d t

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- \frac{2}{e} + 1

= - \frac{2}{e} + 1

\frac{5}{s ( s - 1 )}

x \to 2 lim (\frac{x ^{2} + 3}{x - 2})

\int (1 + 3 x^{2}) ln (2 x) d x

f (x) = 12 x - 2 x

x \frac{d y}{d x} = 2 y

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