What is the integral from 0 to 5 of te^{-t} ?

The integral from 0 to 5 of te^{-t} is -6/(e^5)+1

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integral from 0 to 5 of te^{-t}

\int_{0}^{5} t e^{- t} d t

- \frac{6}{e ^{5}} + 1

Decimal Notation

0.95957 \dots

Solution steps

\int_{0}^{5} t e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t - \int - e^{- t} d t]_{0}^{5}

= [- e^{- t} t - \int - e^{- t} d t]_{0}^{5}

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

= [- e^{- t} t - e^{- t}]_{0}^{5}

Compute the boundaries:

- \frac{6}{e ^{5}} + 1

= - \frac{6}{e ^{5}} + 1

\frac{d}{d x} ((f (x))^{4})

f (x) = sin (6 x)

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

\int_{0}^{1} sin (2 π x) d x

x \to 0 lim (0.9 9^{2} + 1)

What is the integral from 0 to 5 of te^{-t} ?
The integral from 0 to 5 of te^{-t} is -6/(e^5)+1