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

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

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

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

- x e^{- x} - e^{- x} + 1

Solution steps

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

Apply Integration By Parts

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

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

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

Compute the boundaries:

- x e^{- x} - e^{- x} + 1

= - x e^{- x} - e^{- x} + 1

\int_{0}^{1} 12 x^{2} d x

\int_{0}^{5} \frac{8}{7 x + 6} d x

\int_{0}^{10} - x^{2} + 10 x d x

\int_{1}^{4} 2^{2 x} d x

\int_{1}^{2} \frac{x ^{3}}{6} + \frac{1}{2 x} d x

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