What is the integral from 0 to 3 of te^{-t} ?

The integral from 0 to 3 of te^{-t} is -4/(e^3)+1

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integral from 0 to 3 of te^{-t}

\int_{0}^{3} t e^{- t} d t

- \frac{4}{e ^{3}} + 1

Decimal

0.80085 \dots

Solution steps

\int_{0}^{3} t e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t - \int - e^{- t} d t]_{0}^{3}

= [- e^{- t} t - \int - e^{- t} d t]_{0}^{3}

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

= [- e^{- t} t - e^{- t}]_{0}^{3}

Compute the boundaries:

- \frac{4}{e ^{3}} + 1

= - \frac{4}{e ^{3}} + 1

\int 6 x^{2} - 5 x + 3 d x

\int (6 sin (2 x) - 4 cos (4 x)) d x

y^{2} = 4 x, 4 x - 3 y = 4

(arctan (9 x))^{2}

\frac{1}{x ^{5}}

What is the integral from 0 to 3 of te^{-t} ?
The integral from 0 to 3 of te^{-t} is -4/(e^3)+1